From 866032c16d5977a0e55d44b8086c60712032275a Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=97=A8=E5=97=A8=E5=97=A8?= <111850394+QinWeijia111@users.noreply.github.com> Date: Wed, 11 Sep 2024 15:26:40 +0800 Subject: [PATCH 01/64] =?UTF-8?q?Update=200151.=E7=BF=BB=E8=BD=AC=E5=AD=97?= =?UTF-8?q?=E7=AC=A6=E4=B8=B2=E9=87=8C=E7=9A=84=E5=8D=95=E8=AF=8D.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 增加Python语言版本的双指针去除空格算法的非调包算法 --- problems/0151.翻转字符串里的单词.md | 38 +++++++++++++++++++++++++++++ 1 file changed, 38 insertions(+) diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md index a0005198..bf486bdc 100644 --- a/problems/0151.翻转字符串里的单词.md +++ b/problems/0151.翻转字符串里的单词.md @@ -475,7 +475,45 @@ class Solution: words = words[::-1] # 反转单词 return ' '.join(words) #列表转换成字符串 ``` +(版本四) 将字符串转换为列表后，使用双指针去除空格 +```python +class Solution: + def single_reverse(self, s, start: int, end: int): + while start < end: + s[start], s[end] = s[end], s[start] + start += 1 + end -= 1 + def reverseWords(self, s: str) -> str: + result = "" + fast = 0 + # 1. 首先将原字符串反转并且除掉空格, 并且加入到新的字符串当中 + # 由于Python字符串的不可变性，因此只能转换为列表进行处理 + s = list(s) + s.reverse() + while fast < len(s): + if s[fast] != " ": + if len(result) != 0: + result += " " + while s[fast] != " " and fast < len(s): + result += s[fast] + fast += 1 + else: + fast += 1 + # 2.其次将每个单词进行翻转操作 + slow = 0 + fast = 0 + result = list(result) + while fast <= len(result): + if fast == len(result) or result[fast] == " ": + self.single_reverse(result, slow, fast - 1) + slow = fast + 1 + fast += 1 + else: + fast += 1 + + return "".join(result) +``` ### Go： 版本一： From 4f9d9491ae07d68cdd671a28b1b33c294569e00d Mon Sep 17 00:00:00 2001 From: markwang Date: Wed, 11 Sep 2024 17:30:52 +0800 Subject: [PATCH 02/64] =?UTF-8?q?494.=E7=9B=AE=E6=A0=87=E5=92=8C=E5=A2=9E?= =?UTF-8?q?=E5=8A=A0Go=E4=BA=8C=E7=BB=B4dp=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0494.目标和.md | 51 +++++++++++++++++++++++++++++++++++++++++ 1 file changed, 51 insertions(+) diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index 82d330b7..dc8d6c0f 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -675,6 +675,57 @@ class Solution: ``` ### Go +二维dp +```go +func findTargetSumWays(nums []int, target int) int { + sum := 0 + for _, v := range nums { + sum += v + } + if math.Abs(float64(target)) > float64(sum) { + return 0 // 此时没有方案 + } + if (target + sum) % 2 == 1 { + return 0 // 此时没有方案 + } + bagSize := (target + sum) / 2 + + dp := make([][]int, len(nums)) + for i := range dp { + dp[i] = make([]int, bagSize + 1) + } + + // 初始化最上行 + if nums[0] <= bagSize { + dp[0][nums[0]] = 1 + } + + // 初始化最左列，最左列其他数值在递推公式中就完成了赋值 + dp[0][0] = 1 + + var numZero float64 + for i := range nums { + if nums[i] == 0 { + numZero++ + } + dp[i][0] = int(math.Pow(2, numZero)) + } + + // 以下遍历顺序行列可以颠倒 + for i := 1; i < len(nums); i++ { // 行，遍历物品 + for j := 0; j <= bagSize; j++ { // 列，遍历背包 + if nums[i] > j { + dp[i][j] = dp[i-1][j] + } else { + dp[i][j] = dp[i-1][j] + dp[i-1][j-nums[i]] + } + } + } + return dp[len(nums)-1][bagSize] +} +``` + +一维dp ```go func findTargetSumWays(nums []int, target int) int { sum := 0 From b5e3b801d37b290b48d03cdca1ef3a81d680e95b Mon Sep 17 00:00:00 2001 From: Elio Zhou <30508272+eliozh@users.noreply.github.com> Date: Wed, 11 Sep 2024 18:56:36 +0800 Subject: [PATCH 03/64] =?UTF-8?q?Update=200236.=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E6=9C=80=E8=BF=91=E5=85=AC=E5=85=B1=E7=A5=96=E5=85=88?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 优化Rust版本0236.二叉树的最近公共祖先代码 --- problems/0236.二叉树的最近公共祖先.md | 6 +++++- 1 file changed, 5 insertions(+), 1 deletion(-) diff --git a/problems/0236.二叉树的最近公共祖先.md b/problems/0236.二叉树的最近公共祖先.md index 5e80e702..2e94f7a7 100644 --- a/problems/0236.二叉树的最近公共祖先.md +++ b/problems/0236.二叉树的最近公共祖先.md @@ -454,7 +454,11 @@ impl Solution { p: Option>>, q: Option>>, ) -> Option>> { - if root == p || root == q || root.is_none() { + if root.is_none() { + return root; + } + if Rc::ptr_eq(root.as_ref().unwrap(), p.as_ref().unwrap()) + || Rc::ptr_eq(root.as_ref().unwrap(), q.as_ref().unwrap()) { return root; } let left = Self::lowest_common_ancestor( From b7ddc4a4abdb8cde5e020d47d7ed37e18a9df91b Mon Sep 17 00:00:00 2001 From: markwang Date: Thu, 12 Sep 2024 16:07:01 +0800 Subject: [PATCH 04/64] =?UTF-8?q?518.=E9=9B=B6=E9=92=B1=E5=85=91=E6=8D=A2I?= =?UTF-8?q?I=E5=A2=9E=E5=8A=A0Go=E4=BA=8C=E7=BB=B4dp=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0518.零钱兑换II.md | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/problems/0518.零钱兑换II.md b/problems/0518.零钱兑换II.md index 255912d6..bef62b30 100644 --- a/problems/0518.零钱兑换II.md +++ b/problems/0518.零钱兑换II.md @@ -268,6 +268,7 @@ class Solution: ### Go： +一维dp ```go func change(amount int, coins []int) int { // 定义dp数组 @@ -286,6 +287,29 @@ func change(amount int, coins []int) int { return dp[amount] } ``` +二维dp +```go +func change(amount int, coins []int) int { + dp := make([][]int, len(coins)) + for i := range dp { + dp[i] = make([]int, amount + 1) + dp[i][0] = 1 + } + for j := coins[0]; j <= amount; j++ { + dp[0][j] += dp[0][j-coins[0]] + } + for i := 1; i < len(coins); i++ { + for j := 1; j <= amount; j++ { + if j < coins[i] { + dp[i][j] = dp[i-1][j] + } else { + dp[i][j] = dp[i][j-coins[i]] + dp[i-1][j] + } + } + } + return dp[len(coins)-1][amount] +} +``` ### Rust: From b5161e725dc25e2994a903935858d557707cef18 Mon Sep 17 00:00:00 2001 From: markwang Date: Fri, 13 Sep 2024 09:49:36 +0800 Subject: [PATCH 05/64] =?UTF-8?q?kama57.=E7=88=AC=E6=A5=BC=E6=A2=AF?= =?UTF-8?q?=E8=A1=A5=E5=85=85Go=E8=A7=A3=E6=B3=95package=E5=92=8Cimport?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0070.爬楼梯完全背包版本.md | 10 ++++++++++ 1 file changed, 10 insertions(+) diff --git a/problems/0070.爬楼梯完全背包版本.md b/problems/0070.爬楼梯完全背包版本.md index 07e0261e..c51a590b 100644 --- a/problems/0070.爬楼梯完全背包版本.md +++ b/problems/0070.爬楼梯完全背包版本.md @@ -184,6 +184,16 @@ if __name__ == '__main__': ### Go： ```go +package main + +import ( + "bufio" + "fmt" + "os" + "strconv" + "strings" +) + func climbStairs(n int, m int) int { dp := make([]int, n+1) dp[0] = 1 From 8be83e8138532b03aa7b114a7bd84dae325b213c Mon Sep 17 00:00:00 2001 From: cyxiwai Date: Fri, 13 Sep 2024 10:02:13 +0800 Subject: [PATCH 06/64] =?UTF-8?q?=E6=9B=B4=E6=96=B0=E5=B9=B6=E8=A1=A5?= =?UTF-8?q?=E5=85=85=E4=BA=86kama0109.=E5=86=97=E4=BD=99=E9=93=BE=E6=8E=A5?= =?UTF-8?q?II=E7=9A=84Java=E7=89=88=E6=9C=AC=EF=BC=8C=E9=99=84=E6=9C=89?= =?UTF-8?q?=E8=AF=A6=E7=BB=86=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0109.冗余连接II.md | 98 +++++++++++++++++++++++++++ 1 file changed, 98 insertions(+) diff --git a/problems/kamacoder/0109.冗余连接II.md b/problems/kamacoder/0109.冗余连接II.md index fd834357..bd707bf6 100644 --- a/problems/kamacoder/0109.冗余连接II.md +++ b/problems/kamacoder/0109.冗余连接II.md @@ -250,7 +250,105 @@ int main() { ## 其他语言版本 ### Java +```java +import java.util.ArrayList; +import java.util.List; +import java.util.Scanner; +public class Main { + static int n; + static int[] father = new int[1001]; // 并查集数组 + + // 并查集初始化 + public static void init() { + for (int i = 1; i <= n; ++i) { + father[i] = i; + } + } + + // 并查集里寻根的过程 + public static int find(int u) { + if (u == father[u]) return u; + return father[u] = find(father[u]); // 路径压缩 + } + + // 将 v->u 这条边加入并查集 + public static void join(int u, int v) { + u = find(u); + v = find(v); + if (u != v) { + father[v] = u; // 合并两棵树 + } + } + + // 判断 u 和 v 是否有同一个根 + public static boolean same(int u, int v) { + return find(u) == find(v); + } + + // 在有向图里找到删除的那条边，使其变成树 + public static void getRemoveEdge(List edges) { + init(); // 初始化并查集 + for (int i = 0; i < n; i++) { // 遍历所有的边 + if (same(edges.get(i)[0], edges.get(i)[1])) { // 如果构成有向环了，就是要删除的边 + System.out.println(edges.get(i)[0] + " " + edges.get(i)[1]); + return; + } else { + join(edges.get(i)[0], edges.get(i)[1]); + } + } + } + + // 删一条边之后判断是不是树 + public static boolean isTreeAfterRemoveEdge(List edges, int deleteEdge) { + init(); // 初始化并查集 + for (int i = 0; i < n; i++) { + if (i == deleteEdge) continue; + if (same(edges.get(i)[0], edges.get(i)[1])) { // 如果构成有向环了，一定不是树 + return false; + } + join(edges.get(i)[0], edges.get(i)[1]); + } + return true; + } + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + List edges = new ArrayList<>(); // 存储所有的边 + + n = sc.nextInt(); // 顶点数 + int[] inDegree = new int[n + 1]; // 记录每个节点的入度 + for (int i = 0; i < n; i++) { + int s = sc.nextInt(); // 边的起点 + int t = sc.nextInt(); // 边的终点 + inDegree[t]++; + edges.add(new int[]{s, t}); // 将边加入列表 + } + + List vec = new ArrayList<>(); // 记录入度为2的边（如果有的话就两条边） + // 找入度为2的节点所对应的边，注意要倒序，因为优先删除最后出现的一条边 + for (int i = n - 1; i >= 0; i--) { + if (inDegree[edges.get(i)[1]] == 2) { + vec.add(i); + } + } + + // 情况一、情况二 + if (vec.size() > 0) { + // vec里的边已经按照倒叙放的，所以优先删 vec.get(0) 这条边 + if (isTreeAfterRemoveEdge(edges, vec.get(0))) { + System.out.println(edges.get(vec.get(0))[0] + " " + edges.get(vec.get(0))[1]); + } else { + System.out.println(edges.get(vec.get(1))[0] + " " + edges.get(vec.get(1))[1]); + } + return; + } + + // 处理情况三：明确没有入度为2的情况，一定有有向环，找到构成环的边返回即可 + getRemoveEdge(edges); + } +} +``` ### Python ### Go From d6cd5e0ceff6c26dde8083b81b9316b832721bfc Mon Sep 17 00:00:00 2001 From: Yufan SHENG Date: Fri, 13 Sep 2024 14:28:49 +1000 Subject: [PATCH 07/64] =?UTF-8?q?=E6=9B=B4=E6=96=B00110=E5=B9=B3=E8=A1=A1?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91Python3=E8=BF=AD=E4=BB=A3=E6=B3=95?= =?UTF-8?q?=E7=B2=BE=E7=AE=80=E7=89=88=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 9 ++++++--- 1 file changed, 6 insertions(+), 3 deletions(-) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index f8071333..ec120bcc 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -609,10 +609,13 @@ class Solution: while stack: node = stack.pop() if node: - stack.append(node) + stack.append(node) # 中 stack.append(None) - if node.left: stack.append(node.left) - if node.right: stack.append(node.right) + # 采用数组进行迭代，先将右子树加入，保证左节点能够先出栈 + if node.right: # 右 + stack.append(node.right) + if node.left: # 左 + stack.append(node.left) else: real_node = stack.pop() left, right = height_map.get(real_node.left, 0), height_map.get(real_node.right, 0) From 507d9e04308e1859558912a3797e6f9f59106303 Mon Sep 17 00:00:00 2001 From: Yufan Sheng <18829237653@163.com> Date: Fri, 13 Sep 2024 14:40:58 +1000 Subject: [PATCH 08/64] =?UTF-8?q?=E6=9B=B4=E6=96=B00110.=E5=B9=B3=E8=A1=A1?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=E4=B8=AD=E7=9A=84=E4=B8=8A=E6=AC=A1?= =?UTF-8?q?=E4=BF=AE=E6=94=B9=E7=9A=84=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index ec120bcc..a4339ac3 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -611,7 +611,7 @@ class Solution: if node: stack.append(node) # 中 stack.append(None) - # 采用数组进行迭代，先将右子树加入，保证左节点能够先出栈 + # 采用数组进行迭代，先将右节点加入，保证左节点能够先出栈 if node.right: # 右 stack.append(node.right) if node.left: # 左 From 513440ad8f22c6d9a14ceb23a78e0dfde7649c67 Mon Sep 17 00:00:00 2001 From: Chemxy Date: Fri, 13 Sep 2024 19:34:17 +0800 Subject: [PATCH 09/64] =?UTF-8?q?new=20=EF=BC=9A=E6=96=B0=E5=A2=9ECangjie?= =?UTF-8?q?=E8=A7=A3=E9=A2=98=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0300.最长上升子序列.md | 23 +++++++++++++++++++++++ 1 file changed, 23 insertions(+) diff --git a/problems/0300.最长上升子序列.md b/problems/0300.最长上升子序列.md index 9ee7bef3..f1a146b7 100644 --- a/problems/0300.最长上升子序列.md +++ b/problems/0300.最长上升子序列.md @@ -337,6 +337,29 @@ pub fn length_of_lis(nums: Vec) -> i32 { } ``` +### Cangjie: + +```cangjie +func lengthOfLIS(nums: Array): Int64 { + let n = nums.size + if (n <= 1) { + return n + } + + let dp = Array(n, item: 1) + var res = 0 + for (i in 1..n) { + for (j in 0..i) { + if (nums[i] > nums[j]) { + dp[i] = max(dp[i], dp[j] + 1) + } + } + res = max(dp[i], res) + } + return res +} +``` +

From 8127baf3fe23d735a4135900605ad9ebc3ebee63 Mon Sep 17 00:00:00 2001 From: Chemxy Date: Fri, 13 Sep 2024 20:40:37 +0800 Subject: [PATCH 10/64] =?UTF-8?q?new=EF=BC=9A=E6=96=B0=E5=A2=9ECangjie?= =?UTF-8?q?=E9=A2=98=E8=A7=A3=EF=BC=9A=E6=9C=80=E9=95=BF=E8=BF=9E=E7=BB=AD?= =?UTF-8?q?=E9=80=92=E5=A2=9E=E5=BA=8F=E5=88=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0674.最长连续递增序列.md | 19 ++++++++++++++++++- 1 file changed, 18 insertions(+), 1 deletion(-) diff --git a/problems/0674.最长连续递增序列.md b/problems/0674.最长连续递增序列.md index cebb552b..fe882e05 100644 --- a/problems/0674.最长连续递增序列.md +++ b/problems/0674.最长连续递增序列.md @@ -491,7 +491,24 @@ int findLengthOfLCIS(int* nums, int numsSize) { return result; } ``` - +### Cangjie +```cangjie +func findLengthOfLCIS(nums: Array): Int64 { + let n = nums.size + if (n <= 1) { + return n + } + let dp = Array(n, repeat: 1) + var res = 0 + for (i in 1..n) { + if (nums[i] > nums[i - 1]) { + dp[i] = dp[i - 1] + 1 + } + res = max(res, dp[i]) + } + return res +} +``` From 60f34628ebadd556d639ae3792af98df65c3476a Mon Sep 17 00:00:00 2001 From: Chemxy Date: Fri, 13 Sep 2024 20:42:36 +0800 Subject: [PATCH 11/64] =?UTF-8?q?new=EF=BC=9A=E6=96=B0=E5=A2=9ECangjie?= =?UTF-8?q?=E9=A2=98=E8=A7=A3=EF=BC=9A=E6=9C=80=E9=95=BF=E9=87=8D=E5=A4=8D?= =?UTF-8?q?=E5=AD=90=E6=95=B0=E7=BB=84?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0718.最长重复子数组.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/problems/0718.最长重复子数组.md b/problems/0718.最长重复子数组.md index 6c8e7101..19520d13 100644 --- a/problems/0718.最长重复子数组.md +++ b/problems/0718.最长重复子数组.md @@ -581,6 +581,24 @@ int findLength(int* nums1, int nums1Size, int* nums2, int nums2Size) { } ``` +### Cangjie +```cangjie +func findLength(nums1: Array, nums2: Array): Int64 { + let n = nums1.size + let m = nums2.size + let dp = Array(n + 1, {_ => Array(m + 1, item: 0)}) + var res = 0 + for (i in 1..=n) { + for (j in 1..=m) { + if (nums1[i - 1] == nums2[j - 1]) { + dp[i][j] = dp[i - 1][j - 1] + 1 + } + res = max(res, dp[i][j]) + } + } + return res +} +```

From d742497b528e028cff4f756524ed1d9ddc24c8da Mon Sep 17 00:00:00 2001 From: Chemxy Date: Fri, 13 Sep 2024 23:05:37 +0800 Subject: [PATCH 12/64] =?UTF-8?q?new=EF=BC=9A=E6=96=B0=E5=A2=9ECangjie?= =?UTF-8?q?=E9=A2=98=E8=A7=A3=EF=BC=9A=E6=9C=80=E9=95=BF=E5=85=AC=E5=85=B1?= =?UTF-8?q?=E5=AD=90=E5=BA=8F=E5=88=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1143.最长公共子序列.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/problems/1143.最长公共子序列.md b/problems/1143.最长公共子序列.md index 7fa7bb68..93df987e 100644 --- a/problems/1143.最长公共子序列.md +++ b/problems/1143.最长公共子序列.md @@ -399,6 +399,24 @@ int longestCommonSubsequence(char* text1, char* text2) { } ``` +### Cangjie +```cangjie +func longestCommonSubsequence(text1: String, text2: String): Int64 { + let n = text1.size + let m = text2.size + let dp = Array(n + 1, {_ => Array(m + 1, repeat: 0)}) + for (i in 1..=n) { + for (j in 1..=m) { + if (text1[i - 1] == text2[j - 1]) { + dp[i][j] = dp[i - 1][j - 1] + 1 + } else { + dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + } + } + } + return dp[n][m] +} +```

From 8b90ac06bf65ccfa036baa6629a37eb948008108 Mon Sep 17 00:00:00 2001 From: Chemxy Date: Fri, 13 Sep 2024 23:08:46 +0800 Subject: [PATCH 13/64] =?UTF-8?q?fix=EF=BC=9A=E8=B0=83=E6=95=B4=E9=97=B4?= =?UTF-8?q?=E8=B7=9D?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0674.最长连续递增序列.md | 2 ++ problems/0718.最长重复子数组.md | 1 + problems/1143.最长公共子序列.md | 1 + 3 files changed, 4 insertions(+) diff --git a/problems/0674.最长连续递增序列.md b/problems/0674.最长连续递增序列.md index fe882e05..57a38404 100644 --- a/problems/0674.最长连续递增序列.md +++ b/problems/0674.最长连续递增序列.md @@ -491,7 +491,9 @@ int findLengthOfLCIS(int* nums, int numsSize) { return result; } ``` + ### Cangjie + ```cangjie func findLengthOfLCIS(nums: Array): Int64 { let n = nums.size diff --git a/problems/0718.最长重复子数组.md b/problems/0718.最长重复子数组.md index 19520d13..1391926a 100644 --- a/problems/0718.最长重复子数组.md +++ b/problems/0718.最长重复子数组.md @@ -582,6 +582,7 @@ int findLength(int* nums1, int nums1Size, int* nums2, int nums2Size) { ``` ### Cangjie + ```cangjie func findLength(nums1: Array, nums2: Array): Int64 { let n = nums1.size diff --git a/problems/1143.最长公共子序列.md b/problems/1143.最长公共子序列.md index 93df987e..25f32838 100644 --- a/problems/1143.最长公共子序列.md +++ b/problems/1143.最长公共子序列.md @@ -400,6 +400,7 @@ int longestCommonSubsequence(char* text1, char* text2) { ``` ### Cangjie + ```cangjie func longestCommonSubsequence(text1: String, text2: String): Int64 { let n = text1.size From 51f6ab5a54f7e6c3ee0518e2b1fd093b784d434a Mon Sep 17 00:00:00 2001 From: poivre Date: Wed, 18 Sep 2024 00:19:21 +0800 Subject: [PATCH 14/64] =?UTF-8?q?docs:=20=E6=9B=B4=E6=AD=A30459=E7=9A=84?= =?UTF-8?q?=E6=96=87=E6=9C=AC=E7=9A=84=E8=AF=AD=E5=8F=A5=E9=99=88=E8=BF=B0?= =?UTF-8?q?=E3=80=82?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0459.重复的子字符串.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index 254d921d..41c2ea2d 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -163,7 +163,7 @@ KMP算法中next数组为什么遇到字符不匹配的时候可以找到上一 如果一个字符串s是由重复子串组成，那么 最长相等前后缀不包含的子串一定是字符串s的最小重复子串。 -证明： 如果s 是有是有最小重复子串p组成。 +证明： 如果s 是由最小重复子串p组成。 即 s = n * p @@ -884,3 +884,4 @@ public int[] GetNext(string s) + From ee41f4df2203884a884a108bc9cc2662f4fdc823 Mon Sep 17 00:00:00 2001 From: poivre Date: Wed, 18 Sep 2024 00:27:25 +0800 Subject: [PATCH 15/64] =?UTF-8?q?Revert=20"docs:=20=E6=9B=B4=E6=AD=A30459?= =?UTF-8?q?=E7=9A=84=E6=96=87=E6=9C=AC=E7=9A=84=E8=AF=AD=E5=8F=A5=E9=99=88?= =?UTF-8?q?=E8=BF=B0=E3=80=82"?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit This reverts commit 51f6ab5a54f7e6c3ee0518e2b1fd093b784d434a. --- problems/0459.重复的子字符串.md | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index 41c2ea2d..254d921d 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -163,7 +163,7 @@ KMP算法中next数组为什么遇到字符不匹配的时候可以找到上一 如果一个字符串s是由重复子串组成，那么 最长相等前后缀不包含的子串一定是字符串s的最小重复子串。 -证明： 如果s 是由最小重复子串p组成。 +证明： 如果s 是有是有最小重复子串p组成。 即 s = n * p @@ -884,4 +884,3 @@ public int[] GetNext(string s) - From 69a413c3e65ce07144bfdb8199ee0b2d47080e33 Mon Sep 17 00:00:00 2001 From: poivre Date: Wed, 18 Sep 2024 00:28:17 +0800 Subject: [PATCH 16/64] =?UTF-8?q?docs:=20=E6=9B=B4=E6=AD=A30459=E9=97=AE?= =?UTF-8?q?=E9=A2=98=E4=B8=AD=E6=96=87=E6=9C=AC=E7=9A=84=E9=99=88=E8=BF=B0?= =?UTF-8?q?=E3=80=82?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0459.重复的子字符串.md | 5 +++-- 1 file changed, 3 insertions(+), 2 deletions(-) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index 254d921d..dbf32e64 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -163,7 +163,7 @@ KMP算法中next数组为什么遇到字符不匹配的时候可以找到上一 如果一个字符串s是由重复子串组成，那么 最长相等前后缀不包含的子串一定是字符串s的最小重复子串。 -证明： 如果s 是有是有最小重复子串p组成。 +证明： 如果s 是由最小重复子串p组成。 即 s = n * p @@ -179,7 +179,7 @@ KMP算法中next数组为什么遇到字符不匹配的时候可以找到上一 ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20240913110841.png) -这里有录友就想：如果字符串s 是有是有最小重复子串p组成，最长相等前后缀就不能更长一些？ 例如这样： +这里有录友就想：如果字符串s 是由最小重复子串p组成，最长相等前后缀就不能更长一些？ 例如这样： ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20240913114348.png) @@ -884,3 +884,4 @@ public int[] GetNext(string s) + From 75f587507dea0f21ec667a38c049b9324b982f9d Mon Sep 17 00:00:00 2001 From: DraculaJay <113758447+DraculaJay@users.noreply.github.com> Date: Sat, 21 Sep 2024 04:52:40 +0800 Subject: [PATCH 17/64] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E4=BA=86python?= =?UTF-8?q?=E6=96=B0=E7=9A=84=E5=89=AA=E6=9E=9D=E6=93=8D=E4=BD=9C=EF=BC=8C?= =?UTF-8?q?=E9=80=9A=E8=BF=87=E5=AF=B9=E5=89=A9=E4=BD=99=E5=85=83=E7=B4=A0?= =?UTF-8?q?=E9=95=BF=E5=BA=A6=E7=9A=84=E5=88=A4=E6=96=AD=E5=AE=8C=E6=88=90?= =?UTF-8?q?=E5=89=AA=E6=9E=9D?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0093.复原IP地址.md | 32 ++++++++++++++++++++++++++++++-- 1 file changed, 30 insertions(+), 2 deletions(-) diff --git a/problems/0093.复原IP地址.md b/problems/0093.复原IP地址.md index d1300a39..eb81f4b6 100644 --- a/problems/0093.复原IP地址.md +++ b/problems/0093.复原IP地址.md @@ -467,9 +467,37 @@ class Solution: num = int(s[start:end+1]) return 0 <= num <= 255 +回溯（版本三） - - +```python +class Solution: + def restoreIpAddresses(self, s: str) -> List[str]: + result = [] + self.backtracking(s, 0, [], result) + return result + + def backtracking(self, s, startIndex, path, result): + if startIndex == len(s): + result.append('.'.join(path[:])) + return + + for i in range(startIndex, min(startIndex+3, len(s))): + # 如果 i 往后遍历了，并且当前地址的第一个元素是 0 ，就直接退出 + if i > startIndex and s[startIndex] == '0': + break + # 比如 s 长度为 5，当前遍历到 i = 3 这个元素 + # 因为还没有执行任何操作，所以此时剩下的元素数量就是 5 - 3 = 2 ，即包括当前的 i 本身 + # path 里面是当前包含的子串，所以有几个元素就表示储存了几个地址 + # 所以 (4 - len(path)) * 3 表示当前路径至多能存放的元素个数 + # 4 - len(path) 表示至少要存放的元素个数 + if (4 - len(path)) * 3 < len(s) - i or 4 - len(path) > len(s) - i: + break + if i - startIndex == 2: + if not int(s[startIndex:i+1]) <= 255: + break + path.append(s[startIndex:i+1]) + self.backtracking(s, i+1, path, result) + path.pop() ``` ### Go From b3c0d03885496324d5cea3427ccd6cd633368e0a Mon Sep 17 00:00:00 2001 From: Runze Liao <72693579+tony8888lrz@users.noreply.github.com> Date: Fri, 20 Sep 2024 16:56:31 -0700 Subject: [PATCH 18/64] =?UTF-8?q?Update=200343.=E6=95=B4=E6=95=B0=E6=8B=86?= =?UTF-8?q?=E5=88=86.md=20Java=E7=89=88=E6=9C=AC=E5=A2=9E=E5=8A=A0?= =?UTF-8?q?=E8=B4=AA=E5=BF=83=E7=AE=97=E6=B3=95=EF=BC=8C=E6=B8=85=E6=99=B0?= =?UTF-8?q?=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0343.整数拆分.md | 23 +++++++++++++++++++++++ 1 file changed, 23 insertions(+) diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md index 7295627f..3ba23e52 100644 --- a/problems/0343.整数拆分.md +++ b/problems/0343.整数拆分.md @@ -243,6 +243,29 @@ class Solution { } } ``` +贪心 +```Java +class Solution { + public int integerBreak(int n) { + // with 贪心 + // 通过数学原理拆出更多的3乘积越大，则 + /** + @Param: an int, the integer we need to break. + @Return: an int, the maximum integer after breaking + @Method: Using math principle to solve this problem + @Time complexity: O(1) + **/ + if(n == 2) return 1; + if(n == 3) return 2; + int result = 1; + while(n > 4) { + n-=3; + result *=3; + } + return result*n; + } +} +``` ### Python 动态规划（版本一） From fed25bf0125118e4b985fb58d122430b3141b919 Mon Sep 17 00:00:00 2001 From: suinming <0223314338aa@gmail.com> Date: Sat, 21 Sep 2024 14:36:45 +0800 Subject: [PATCH 19/64] =?UTF-8?q?feat:=20=E5=9C=96=E8=AB=96100=E5=B2=9B?= =?UTF-8?q?=E5=B1=BF=E7=9A=84=E6=9C=80=E5=A4=A7=E9=9D=A2=E7=A7=AF=EF=BC=8C?= =?UTF-8?q?=E6=96=B0=E5=A2=9Ejs=E6=B7=B1=E6=90=9C=E7=AE=97=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0100.岛屿的最大面积.md | 77 +++++++++++++++++++++++ 1 file changed, 77 insertions(+) diff --git a/problems/kamacoder/0100.岛屿的最大面积.md b/problems/kamacoder/0100.岛屿的最大面积.md index ea62edc2..51bfc57f 100644 --- a/problems/kamacoder/0100.岛屿的最大面积.md +++ b/problems/kamacoder/0100.岛屿的最大面积.md @@ -480,7 +480,84 @@ const bfs = (graph, visited, x, y) => { })() ``` +```javascript +// 深搜版 + +const r1 = require('readline').createInterface({ input: process.stdin }); +// 创建readline接口 +let iter = r1[Symbol.asyncIterator](); +// 创建异步迭代器 +const readline = async () => (await iter.next()).value; + +let graph // 地图 +let N, M // 地图大小 +let visited // 访问过的节点 +let result = 0 // 最大岛屿面积 +let count = 0 // 岛屿内节点数 +const dir = [[0, 1], [1, 0], [0, -1], [-1, 0]] //方向 + +// 读取输入，初始化地图 +const initGraph = async () => { + let line = await readline(); + [N, M] = line.split(' ').map(Number); + graph = new Array(N).fill(0).map(() => new Array(M).fill(0)) + visited = new Array(N).fill(false).map(() => new Array(M).fill(false)) + + for (let i = 0; i < N; i++) { + line = await readline() + line = line.split(' ').map(Number) + for (let j = 0; j < M; j++) { + graph[i][j] = line[j] + } + } +} + +/** + * @description: 从(x, y)开始深度优先遍历 + * @param {*} graph 地图 + * @param {*} visited 访问过的节点 + * @param {*} x 开始搜索节点的下标 + * @param {*} y 开始搜索节点的下标 + * @return {*} + */ +const dfs = (graph, visited, x, y) => { + for (let i = 0; i < 4; i++) { + let nextx = x + dir[i][0] + let nexty = y + dir[i][1] + if(nextx < 0 || nextx >= N || nexty < 0 || nexty >= M) continue + if(!visited[nextx][nexty] && graph[nextx][nexty] === 1){ + count++ + visited[nextx][nexty] = true + dfs(graph, visited, nextx, nexty) + } + } +} + +(async function () { + + // 读取输入，初始化地图 + await initGraph() + + // 统计最大岛屿面积 + for (let i = 0; i < N; i++) { + for (let j = 0; j < M; j++) { + if (!visited[i][j] && graph[i][j] === 1) { //遇到没有访问过的陆地 + // 重新计算面积 + count = 1 + visited[i][j] = true + + // 深度优先遍历，统计岛屿内节点数，并将岛屿标记为已访问 + dfs(graph, visited, i, j) + + // 更新最大岛屿面积 + result = Math.max(result, count) + } + } + } + console.log(result); +})() +``` ### TypeScript From 99e85bfea1804fcc87458d858fed44d76c7996f2 Mon Sep 17 00:00:00 2001 From: suinming <0223314338aa@gmail.com> Date: Sat, 21 Sep 2024 16:05:46 +0800 Subject: [PATCH 20/64] =?UTF-8?q?feat:=20=E5=9C=96=E8=AB=96101=E5=AD=A4?= =?UTF-8?q?=E5=B3=B6=E6=80=BB=E9=9D=A2=E7=A7=AF=20=EF=BC=8C=E6=96=B0?= =?UTF-8?q?=E5=A2=9Epython=E6=B7=B1=E6=90=9C=E7=AE=97=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0101.孤岛的总面积.md | 65 +++++++++++++++++++++++++ 1 file changed, 65 insertions(+) diff --git a/problems/kamacoder/0101.孤岛的总面积.md b/problems/kamacoder/0101.孤岛的总面积.md index 006484de..26c92c07 100644 --- a/problems/kamacoder/0101.孤岛的总面积.md +++ b/problems/kamacoder/0101.孤岛的总面积.md @@ -307,6 +307,71 @@ for i in range(n): print(count) ``` + +```python +direction = [[1, 0], [-1, 0], [0, 1], [0, -1]] +result = 0 + +# 深度搜尋 +def dfs(grid, y, x): + grid[y][x] = 0 + global result + result += 1 + + for i, j in direction: + next_x = x + j + next_y = y + i + if (next_x < 0 or next_y < 0 or + next_x >= len(grid[0]) or next_y >= len(grid) + ): + continue + if grid[next_y][next_x] == 1 and not visited[next_y][next_x]: + visited[next_y][next_x] = True + dfs(grid, next_y, next_x) + + +# 讀取輸入值 +n, m = map(int, input().split()) +grid = [] +visited = [[False] * m for _ in range(n)] + +for i in range(n): + grid.append(list(map(int, input().split()))) + +# 處理邊界 +for j in range(m): + # 上邊界 + if grid[0][j] == 1 and not visited[0][j]: + visited[0][j] = True + dfs(grid, 0, j) + # 下邊界 + if grid[n - 1][j] == 1 and not visited[n - 1][j]: + visited[n - 1][j] = True + dfs(grid, n - 1, j) + +for i in range(n): + # 左邊界 + if grid[i][0] == 1 and not visited[i][0]: + visited[i][0] = True + dfs(grid, i, 0) + # 右邊界 + if grid[i][m - 1] == 1 and not visited[i][m - 1]: + visited[i][m - 1] = True + dfs(grid, i, m - 1) + +# 計算孤島總面積 +result = 0 # 初始化，避免使用到處理邊界時所產生的累加值 + +for i in range(n): + for j in range(m): + if grid[i][j] == 1 and not visited[i][j]: + visited[i][j] = True + dfs(grid, i, j) + +# 輸出孤島的總面積 +print(result) +``` + ### Go ``` go From 848fbd4cd5c868839c75c637ef5ea9057ab301d7 Mon Sep 17 00:00:00 2001 From: Zhihan Li <54661071+zhihali@users.noreply.github.com> Date: Sun, 22 Sep 2024 16:22:04 +0100 Subject: [PATCH 21/64] =?UTF-8?q?Update=200053.=E6=9C=80=E5=A4=A7=E5=AD=90?= =?UTF-8?q?=E5=BA=8F=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0053.最大子序和.md | 15 +++++++++++++-- 1 file changed, 13 insertions(+), 2 deletions(-) diff --git a/problems/0053.最大子序和.md b/problems/0053.最大子序和.md index 551c39bf..1c7ff0cd 100644 --- a/problems/0053.最大子序和.md +++ b/problems/0053.最大子序和.md @@ -214,6 +214,7 @@ class Solution: return result ``` +贪心法 ```python class Solution: def maxSubArray(self, nums): @@ -226,8 +227,18 @@ class Solution: if count <= 0: # 相当于重置最大子序起始位置，因为遇到负数一定是拉低总和 count = 0 return result - - +``` +动态规划 +```python +class Solution: + def maxSubArray(self, nums: List[int]) -> int: + dp = [0] * len(nums) + dp[0] = nums[0] + res = nums[0] + for i in range(1, len(nums)): + dp[i] = max(dp[i-1] + nums[i], nums[i]) + res = max(res, dp[i]) + return res ``` ### Go 贪心法 From 2841f59212a4f8399d53007eb01130172ce59814 Mon Sep 17 00:00:00 2001 From: suinming <0223314338aa@gmail.com> Date: Tue, 24 Sep 2024 14:41:38 +0800 Subject: [PATCH 22/64] =?UTF-8?q?feat:=20108.=20=E5=86=97=E4=BD=99?= =?UTF-8?q?=E8=BF=9E=E6=8E=A5=E6=96=B0=E5=A2=9Epython=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0108.冗余连接.md | 39 +++++++++++++++++++++++++++++ 1 file changed, 39 insertions(+) diff --git a/problems/kamacoder/0108.冗余连接.md b/problems/kamacoder/0108.冗余连接.md index 2c133782..18a86ad6 100644 --- a/problems/kamacoder/0108.冗余连接.md +++ b/problems/kamacoder/0108.冗余连接.md @@ -178,6 +178,45 @@ int main() { ### Python +```python +father = list() + +def find(u): + if u == father[u]: + return u + else: + father[u] = find(father[u]) + return father[u] + +def is_same(u, v): + u = find(u) + v = find(v) + return u == v + +def join(u, v): + u = find(u) + v = find(v) + if u != v: + father[u] = v + +if __name__ == "__main__": + # 輸入 + n = int(input()) + for i in range(n + 1): + father.append(i) + # 尋找冗余邊 + result = None + for i in range(n): + s, t = map(int, input().split()) + if is_same(s, t): + result = str(s) + ' ' + str(t) + else: + join(s, t) + + # 輸出 + print(result) +``` + ### Go ### Rust From 5c67ef4acd04b228b2f24f32a89917d9abc478b0 Mon Sep 17 00:00:00 2001 From: suinming <0223314338aa@gmail.com> Date: Tue, 24 Sep 2024 16:13:28 +0800 Subject: [PATCH 23/64] =?UTF-8?q?feat:=20109.=20=E5=86=97=E4=BD=99?= =?UTF-8?q?=E8=BF=9E=E6=8E=A5II=E6=96=B0=E5=A2=9Epython=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0109.冗余连接II.md | 86 +++++++++++++++++++++++++++ 1 file changed, 86 insertions(+) diff --git a/problems/kamacoder/0109.冗余连接II.md b/problems/kamacoder/0109.冗余连接II.md index bd707bf6..2bd4eac6 100644 --- a/problems/kamacoder/0109.冗余连接II.md +++ b/problems/kamacoder/0109.冗余连接II.md @@ -351,6 +351,92 @@ public class Main { ``` ### Python +```python +from collections import defaultdict + +father = list() + + +def find(u): + if u == father[u]: + return u + else: + father[u] = find(father[u]) + return father[u] + + +def is_same(u, v): + u = find(u) + v = find(v) + return u == v + + +def join(u, v): + u = find(u) + v = find(v) + if u != v: + father[u] = v + + +def is_tree_after_remove_edge(edges, edge, n): + # 初始化并查集 + global father + father = [i for i in range(n + 1)] + + for i in range(len(edges)): + if i == edge: + continue + s, t = edges[i] + if is_same(s, t): # 成環，即不是有向樹 + return False + else: # 將s,t放入集合中 + join(s, t) + return True + + +def get_remove_edge(edges): + # 初始化并查集 + global father + father = [i for i in range(n + 1)] + + for s, t in edges: + if is_same(s, t): + print(s, t) + return + else: + join(s, t) + + +if __name__ == "__main__": + # 輸入 + n = int(input()) + edges = list() + in_degree = defaultdict(int) + + for i in range(n): + s, t = map(int, input().split()) + in_degree[t] += 1 + edges.append([s, t]) + + # 尋找入度為2的邊，並紀錄其下標(index) + vec = list() + for i in range(n - 1, -1, -1): + if in_degree[edges[i][1]] == 2: + vec.append(i) + + # 輸出 + if len(vec) > 0: + # 情況一：刪除輸出順序靠後的邊 + if is_tree_after_remove_edge(edges, vec[0], n): + print(edges[vec[0]][0], edges[vec[0]][1]) + # 情況二：只能刪除特定的邊 + else: + print(edges[vec[1]][0], edges[vec[1]][1]) + else: + # 情況三： 原圖有環 + get_remove_edge(edges) +``` + ### Go ### Rust From 183fe44ae0d0569f39ab525b57877d844e9d3823 Mon Sep 17 00:00:00 2001 From: Zhihan Li <54661071+zhihali@users.noreply.github.com> Date: Tue, 24 Sep 2024 16:09:20 +0100 Subject: [PATCH 24/64] =?UTF-8?q?Update=200134.=E5=8A=A0=E6=B2=B9=E7=AB=99?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0134.加油站.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0134.加油站.md b/problems/0134.加油站.md index 7ac9f0f9..5cf50b3e 100644 --- a/problems/0134.加油站.md +++ b/problems/0134.加油站.md @@ -158,7 +158,7 @@ i从0开始累加rest[i]，和记为curSum，一旦curSum小于零，说明[0, i 如果 curSum<0 说明 区间和1 + 区间和2 < 0， 那么 假设从上图中的位置开始计数curSum不会小于0的话，就是 区间和2>0。 -区间和1 + 区间和2 < 0 同时 区间和2>0，只能说明区间和1 < 0， 那么就会从假设的箭头初就开始从新选择其实位置了。 +区间和1 + 区间和2 < 0 同时 区间和2>0，只能说明区间和1 < 0， 那么就会从假设的箭头初就开始从新选择起始位置了。 **那么局部最优：当前累加rest[i]的和curSum一旦小于0，起始位置至少要是i+1，因为从i之前开始一定不行。全局最优：找到可以跑一圈的起始位置**。 From 779ff9dd08a4b18d0aa8e077fcdab5e821662c08 Mon Sep 17 00:00:00 2001 From: suinming <0223314338aa@gmail.com> Date: Wed, 25 Sep 2024 12:35:34 +0800 Subject: [PATCH 25/64] =?UTF-8?q?feat:=2053.=20=E5=AF=BB=E5=AE=9D=E6=96=B0?= =?UTF-8?q?=E5=A2=9Ejs=E7=AE=97=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0053.寻宝-prim.md | 49 ++++++++++++++++++++++++++++ 1 file changed, 49 insertions(+) diff --git a/problems/kamacoder/0053.寻宝-prim.md b/problems/kamacoder/0053.寻宝-prim.md index c71624b5..a8dad4cb 100644 --- a/problems/kamacoder/0053.寻宝-prim.md +++ b/problems/kamacoder/0053.寻宝-prim.md @@ -693,6 +693,55 @@ if __name__ == "__main__": ### Rust ### Javascript +```js +function prim(v, edges) { + const grid = Array.from({ length: v + 1 }, () => new Array(v + 1).fill(10001)); // Fixed grid initialization + const minDist = new Array(v + 1).fill(10001) + const isInTree = new Array(v + 1).fill(false) + // 建構鄰接矩陣 + for(const [v1, v2, w] of edges) { + grid[v1][v2] = w + grid[v2][v1] = w + } + // prim 演算法 + for (let i = 1 ; i < v ; i++) { + let cur = -1 + let tempMinDist = Number.MAX_VALUE + // 1. 尋找距離生成樹最近的節點 + for (let j = 1 ; j < v + 1 ; j++) { + if (!isInTree[j] && minDist[j] < tempMinDist) { + tempMinDist = minDist[j] + cur = j + } + } + // 2. 將節點放入生成樹 + isInTree[cur] = true + // 3. 更新非生成樹節點與生成樹的最短距離 + for (let j = 1 ; j < v + 1 ; j++) { + if (!isInTree[j] && grid[cur][j] < minDist[j]) { + minDist[j] = grid[cur][j] + } + } + } + console.log(minDist.slice(2).reduce((acc, cur) => acc + cur, 0)) +} + + +async function main() { + const rl = require('readline').createInterface({ input: process.stdin }) + const iter = rl[Symbol.asyncIterator]() + const readline = async () => (await iter.next()).value + const [v, e] = (await readline()).split(" ").map(Number) + const edges = [] + for (let i = 0 ; i < e ; i++) { + edges.push((await readline()).split(" ").map(Number)) + } + prim(v, edges) +} + + +main() +``` ### TypeScript From 9e66f2232363c66f1c466329b518fc2d8c6525c5 Mon Sep 17 00:00:00 2001 From: suinming <0223314338aa@gmail.com> Date: Wed, 25 Sep 2024 14:57:21 +0800 Subject: [PATCH 26/64] =?UTF-8?q?feat:=2053.=20=E5=AF=BB=E5=AE=9D=E6=96=B0?= =?UTF-8?q?=E5=A2=9Ekruskal=20js=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0053.寻宝-Kruskal.md | 56 +++++++++++++++++++++++++ 1 file changed, 56 insertions(+) diff --git a/problems/kamacoder/0053.寻宝-Kruskal.md b/problems/kamacoder/0053.寻宝-Kruskal.md index cb24fd17..6a227985 100644 --- a/problems/kamacoder/0053.寻宝-Kruskal.md +++ b/problems/kamacoder/0053.寻宝-Kruskal.md @@ -549,6 +549,62 @@ if __name__ == "__main__": ### Javascript +```js +function kruskal(v, edges) { + const father = Array.from({ length: v + 1 }, (_, i) => i) + + function find(u){ + if (u === father[u]) { + return u + } else { + father[u] = find(father[u]) + return father[u] + } + + } + + function isSame(u, v) { + let s = find(u) + let t = find(v) + return s === t + } + + function join(u, v) { + let s = find(u) + let t = find(v) + if (s !== t) { + father[s] = t + } + } + + edges.sort((a, b) => a[2] - b[2]) + let result = 0 + for (const [v1, v2, w] of edges) { + if (!isSame(v1, v2)) { + result += w + join(v1 ,v2) + } + } + console.log(result) +} + + +async function main() { + const rl = require('readline').createInterface({ input: process.stdin }) + const iter = rl[Symbol.asyncIterator]() + const readline = async () => (await iter.next()).value + const [v, e] = (await readline()).split(" ").map(Number) + const edges = [] + for (let i = 0 ; i < e ; i++) { + edges.push((await readline()).split(" ").map(Number)) + } + kruskal(v, edges) +} + + +main() +``` + ### TypeScript ### PhP From 13a4234fce501cf300974d4f28f7728c0f91e64e Mon Sep 17 00:00:00 2001 From: xiaodaoshou <1035011225@qq.com> Date: Wed, 25 Sep 2024 20:50:05 +0800 Subject: [PATCH 27/64] =?UTF-8?q?0005=E6=9C=80=E9=95=BF=E5=9B=9E=E6=96=87?= =?UTF-8?q?=E5=AD=90=E4=B8=B2=20Manacher=E7=AE=97=E6=B3=95C++=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0005.最长回文子串.md | 54 +++++++++++++++++++++++++++++++++++ 1 file changed, 54 insertions(+) diff --git a/problems/0005.最长回文子串.md b/problems/0005.最长回文子串.md index a13daf1e..b3d3b938 100644 --- a/problems/0005.最长回文子串.md +++ b/problems/0005.最长回文子串.md @@ -256,7 +256,60 @@ public: * 时间复杂度：O(n^2) * 空间复杂度：O(1) +### Manacher 算法 +Manacher 算法的关键在于高效利用回文的对称性，通过插入分隔符和维护中心、边界等信息，在线性时间内找到最长回文子串。这种方法避免了重复计算，是处理回文问题的最优解。 + +```c++ +//Manacher 算法 +class Solution { +public: + string longestPalindrome(string s) { + // 预处理字符串，在每个字符之间插入 '#' + string t = "#"; + for (char c : s) { + t += c; // 添加字符 + t += '#';// 添加分隔符 + } + int n = t.size();// 新字符串的长度 + vector p(n, 0);// p[i] 表示以 t[i] 为中心的回文半径 + int center = 0, right = 0;// 当前回文的中心和右边界 + + + // 遍历预处理后的字符串 + for (int i = 0; i < n; i++) { + // 如果当前索引在右边界内，利用对称性初始化 p[i] + if (i < right) { + p[i] = min(right - i, p[2 * center - i]); + } + // 尝试扩展回文 + while (i - p[i] - 1 >= 0 && i + p[i] + 1 < n && t[i - p[i] - 1] == t[i + p[i] + 1]) { + p[i]++;// 增加回文半径 + } + // 如果当前回文扩展超出右边界，更新中心和右边界 + if (i + p[i] > right) { + center = i;// 更新中心 + right = i + p[i];// 更新右边界 + } + } + // 找到最大回文半径和对应的中心 + int maxLen = 0, centerIndex = 0; + for (int i = 0; i < n; i++) { + if (p[i] > maxLen) { + maxLen = p[i];// 更新最大回文长度 + centerIndex = i;// 更新中心索引 + } + } + // 计算原字符串中回文子串的起始位置并返回 + return s.substr((centerIndex - maxLen) / 2, maxLen); + } +}; +``` + + + +* 时间复杂度：O(n) +* 空间复杂度：O(n) ## 其他语言版本 @@ -682,3 +735,4 @@ public class Solution { + From 1b08a8e5b5b1cb83ca55f2d42391791387f293dd Mon Sep 17 00:00:00 2001 From: suinming <0223314338aa@gmail.com> Date: Thu, 26 Sep 2024 12:56:42 +0800 Subject: [PATCH 28/64] =?UTF-8?q?feat:=2047.=20=E5=8F=82=E5=8A=A0=E7=A7=91?= =?UTF-8?q?=E5=AD=A6=E5=A4=A7=E4=BC=9A=E6=96=B0=E5=A2=9Ejs=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0047.参会dijkstra朴素.md | 59 +++++++++++++++++++++ 1 file changed, 59 insertions(+) diff --git a/problems/kamacoder/0047.参会dijkstra朴素.md b/problems/kamacoder/0047.参会dijkstra朴素.md index c0a490b3..465ad16d 100644 --- a/problems/kamacoder/0047.参会dijkstra朴素.md +++ b/problems/kamacoder/0047.参会dijkstra朴素.md @@ -869,6 +869,65 @@ if __name__ == "__main__": ### Javascript +```js +function dijkstra(grid, start, end) { + const visited = Array.from({length: end + 1}, () => false) + const minDist = Array.from({length: end + 1}, () => Number.MAX_VALUE) + minDist[start] = 0 + + for (let i = 1 ; i < end + 1 ; i++) { + let cur = -1 + let tempMinDist = Number.MAX_VALUE + // 1. 找尋與起始點距離最近且未被訪的節點 + for (let j = 1 ; j < end + 1 ; j++) { + if (!visited[j] && minDist[j] < tempMinDist) { + cur = j + tempMinDist = minDist[j] + } + } + if (cur === -1) break; + + // 2. 更新節點狀態為已拜訪 + visited[cur] = true + + // 3. 更新未拜訪節點與起始點的最短距離 + for (let j = 1 ; j < end + 1 ; j++) { + if(!visited[j] && grid[cur][j] != Number.MAX_VALUE + && grid[cur][j] + minDist[cur] < minDist[j] + ) { + minDist[j] = grid[cur][j] + minDist[cur] + } + } + } + + return minDist[end] === Number.MAX_VALUE ? -1 : minDist[end] +} + + +async function main() { + // 輸入 + const rl = require('readline').createInterface({ input: process.stdin }) + const iter = rl[Symbol.asyncIterator]() + const readline = async () => (await iter.next()).value + const [n, m] = (await readline()).split(" ").map(Number) + const grid = Array.from({length: n + 1}, + () => Array.from({length:n + 1}, () => Number.MAX_VALUE)) + for (let i = 0 ; i < m ; i++) { + const [s, e, w] = (await readline()).split(" ").map(Number) + grid[s][e] = w + } + + // dijkstra + const result = dijkstra(grid, 1, n) + + // 輸出 + console.log(result) +} + + +main() +``` + ### TypeScript ### PhP From 53a4a17b8c518b6e27166b3b9b60de44fdf9b39a Mon Sep 17 00:00:00 2001 From: Camille <59353274+Camille0512@users.noreply.github.com> Date: Thu, 26 Sep 2024 23:06:45 +0800 Subject: [PATCH 29/64] =?UTF-8?q?Update=200099.=E5=B2=9B=E5=B1=BF=E7=9A=84?= =?UTF-8?q?=E6=95=B0=E9=87=8F=E5=B9=BF=E6=90=9C.md:=20Add=20Scala=20script?= =?UTF-8?q?=20for=20the=20counting=20island=20puzzle=20using=20BFS=20algo?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0099.岛屿的数量广搜.md | 49 +++++++++++++++++++++++ 1 file changed, 49 insertions(+) diff --git a/problems/kamacoder/0099.岛屿的数量广搜.md b/problems/kamacoder/0099.岛屿的数量广搜.md index 30475753..9d31c922 100644 --- a/problems/kamacoder/0099.岛屿的数量广搜.md +++ b/problems/kamacoder/0099.岛屿的数量广搜.md @@ -499,6 +499,55 @@ main(); ### Swift ### Scala +```scala +import scala.collection.mutable.Queue +import util.control.Breaks._ + +// Dev on LeetCode: https://leetcode.cn/problems/number-of-islands/description/ +object Solution { + def numIslands(grid: Array[Array[Char]]): Int = { + val row = grid.length + val col = grid(0).length + val dir = List((-1,0), (0,-1), (1,0), (0,1)) // 四个方向 + var visited = Array.fill(row)(Array.fill(col)(false)) + var counter = 0 + var que = Queue.empty[Tuple2[Int, Int]] + + (0 until row).map{ r => + (0 until col).map{ c => + breakable { + if (!visited(r)(c) && grid(r)(c) == '1') { + que.enqueue((r, c)) + visited(r)(c) // 只要加入队列，立刻标记 + } else break // 不是岛屿不进入queue，也不记录 + + while (!que.isEmpty) { + val cur = que.head + que.dequeue() + val x = cur(0) + val y = cur(1) + dir.map{ d => + val nextX = x + d(0) + val nextY = y + d(1) + breakable { + // 越界就跳过 + if (nextX < 0 || nextX >= row || nextY < 0 || nextY >= col) break + if (!visited(nextX)(nextY) && grid(nextX)(nextY) == '1') { + visited(nextX)(nextY) = true // 只要加入队列，立刻标记 + que.enqueue((nextX, nextY)) + } + } + } + } + counter = counter + 1 // 找完一个岛屿后记录一下 + } + } + } + + counter + } +} +``` ### C# From 8ff94842609eee048ec78b709a20a82339c3c093 Mon Sep 17 00:00:00 2001 From: suinming <0223314338aa@gmail.com> Date: Fri, 27 Sep 2024 15:20:59 +0800 Subject: [PATCH 30/64] =?UTF-8?q?feat:=2094.=20=E5=9F=8E=E5=B8=82=E9=97=B4?= =?UTF-8?q?=E8=B4=A7=E7=89=A9=E8=BF=90=E8=BE=93=20I=20=E6=96=B0=E5=A2=9Ejs?= =?UTF-8?q?=E7=AE=97=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0094.城市间货物运输I.md | 39 ++++++++++++++++++++++ 1 file changed, 39 insertions(+) diff --git a/problems/kamacoder/0094.城市间货物运输I.md b/problems/kamacoder/0094.城市间货物运输I.md index 3737fe01..9021e0fe 100644 --- a/problems/kamacoder/0094.城市间货物运输I.md +++ b/problems/kamacoder/0094.城市间货物运输I.md @@ -485,6 +485,45 @@ if __name__ == "__main__": ### Javascript +```js +async function main() { + // 輸入 + const rl = require('readline').createInterface({ input: process.stdin }) + const iter = rl[Symbol.asyncIterator]() + const readline = async () => (await iter.next()).value + const [n, m] = (await readline()).split(" ").map(Number) + const edges = [] + for (let i = 0 ; i < m ; i++) { + edges.push((await readline()).split(" ").map(Number)) + } + const minDist = Array.from({length: n + 1}, () => Number.MAX_VALUE) + // 起始點 + minDist[1] = 0 + + for (let i = 1 ; i < n ; i++) { + let update = false + for (const [src, desc, w] of edges) { + if (minDist[src] !== Number.MAX_VALUE && minDist[src] + w < minDist[desc]) { + minDist[desc] = minDist[src] + w + update = true + } + } + if (!update) { + break; + } + } + + // 輸出 + if (minDist[n] === Number.MAX_VALUE) { + console.log('unconnected') + } else { + console.log(minDist[n]) + } +} + +main() +``` + ### TypeScript ### PhP From c55a05a496fdb75c1f38fe49a060fa3985565342 Mon Sep 17 00:00:00 2001 From: suinming <0223314338aa@gmail.com> Date: Fri, 27 Sep 2024 15:59:22 +0800 Subject: [PATCH 31/64] =?UTF-8?q?feat:=20=2094.=20=E5=9F=8E=E5=B8=82?= =?UTF-8?q?=E9=97=B4=E8=B4=A7=E7=89=A9=E8=BF=90=E8=BE=93=20I=20SPEA?= =?UTF-8?q?=E6=96=B0=E5=A2=9Ejs=E7=AE=97=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../kamacoder/0094.城市间货物运输I-SPFA.md | 54 +++++++++++++++++++ 1 file changed, 54 insertions(+) diff --git a/problems/kamacoder/0094.城市间货物运输I-SPFA.md b/problems/kamacoder/0094.城市间货物运输I-SPFA.md index b3f42bf8..9ba92599 100644 --- a/problems/kamacoder/0094.城市间货物运输I-SPFA.md +++ b/problems/kamacoder/0094.城市间货物运输I-SPFA.md @@ -464,6 +464,60 @@ if __name__ == "__main__": ### Javascript +```js +async function main() { + // 輸入 + const rl = require('readline').createInterface({ input: process.stdin }) + const iter = rl[Symbol.asyncIterator]() + const readline = async () => (await iter.next()).value + const [n, m] = (await readline()).split(" ").map(Number) + const grid = {} + for (let i = 0 ; i < m ; i++) { + const [src, desc, w] = (await readline()).split(" ").map(Number) + if (grid.hasOwnProperty(src)) { + grid[src].push([desc, w]) + } else { + grid[src] = [[desc, w]] + } + } + const minDist = Array.from({length: n + 1}, () => Number.MAX_VALUE) + + // 起始點 + minDist[1] = 0 + + const q = [1] + const visited = Array.from({length: n + 1}, () => false) + + while (q.length) { + const src = q.shift() + const neighbors = grid[src] + visited[src] = false + if (neighbors) { + for (const [desc, w] of neighbors) { + if (minDist[src] !== Number.MAX_VALUE + && minDist[src] + w < minDist[desc]) { + minDist[desc] = minDist[src] + w + if (!visited[desc]) { + q.push(desc) + visited[desc] = true + } + + } + } + } + } + + // 輸出 + if (minDist[n] === Number.MAX_VALUE) { + console.log('unconnected') + } else { + console.log(minDist[n]) + } +} + +main() +``` + ### TypeScript ### PhP From ecf70c6e57dc5d6e7583f9d362d5420f2d6dc74a Mon Sep 17 00:00:00 2001 From: suinming <0223314338aa@gmail.com> Date: Fri, 27 Sep 2024 19:43:51 +0800 Subject: [PATCH 32/64] =?UTF-8?q?feat:=2096.=20=E5=9F=8E=E5=B8=82=E9=97=B4?= =?UTF-8?q?=E8=B4=A7=E7=89=A9=E8=BF=90=E8=BE=93=20III=E5=A2=9E=E5=8A=A0pyt?= =?UTF-8?q?hon=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0096.城市间货物运输III.md | 36 ++++++++++++++++++++ 1 file changed, 36 insertions(+) diff --git a/problems/kamacoder/0096.城市间货物运输III.md b/problems/kamacoder/0096.城市间货物运输III.md index dacd23d1..567a1d87 100644 --- a/problems/kamacoder/0096.城市间货物运输III.md +++ b/problems/kamacoder/0096.城市间货物运输III.md @@ -703,6 +703,42 @@ public class Main { ``` ### Python +```python +def main(): + # 輸入 + n, m = map(int, input().split()) + edges = list() + for _ in range(m): + edges.append(list(map(int, input().split() ))) + + start, end, k = map(int, input().split()) + min_dist = [float('inf') for _ in range(n + 1)] + min_dist[start] = 0 + + # 只能經過k個城市，所以從起始點到中間有(k + 1)個邊連接 + # 需要鬆弛(k + 1)次 + + for _ in range(k + 1): + update = False + min_dist_copy = min_dist.copy() + for src, desc, w in edges: + if (min_dist_copy[src] != float('inf') and + min_dist_copy[src] + w < min_dist[desc]): + min_dist[desc] = min_dist_copy[src] + w + update = True + if not update: + break + # 輸出 + if min_dist[end] == float('inf'): + print('unreachable') + else: + print(min_dist[end]) + + + +if __name__ == "__main__": + main() +``` ### Go From b32888260f0cb2f29314cf0c4d7d4dddb1f20f71 Mon Sep 17 00:00:00 2001 From: Ziyang Wen Date: Fri, 27 Sep 2024 21:17:44 +0800 Subject: [PATCH 33/64] =?UTF-8?q?0151python=E7=89=88=E6=9C=AC1=E5=88=A0?= =?UTF-8?q?=E9=99=A4=E5=86=97=E4=BD=99=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0151.翻转字符串里的单词.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md index bf486bdc..9a0cbea4 100644 --- a/problems/0151.翻转字符串里的单词.md +++ b/problems/0151.翻转字符串里的单词.md @@ -440,11 +440,10 @@ class Solution { ```Python class Solution: def reverseWords(self, s: str) -> str: - # 删除前后空白 - s = s.strip() # 反转整个字符串 s = s[::-1] # 将字符串拆分为单词，并反转每个单词 + # split()函数能够自动忽略多余的空白字符 s = ' '.join(word[::-1] for word in s.split()) return s @@ -1029,3 +1028,4 @@ public string ReverseWords(string s) { + From 0c29c3ed1edd5fcf5be9a8eb6dcba6fc25fb569b Mon Sep 17 00:00:00 2001 From: Ziyang Wen Date: Fri, 27 Sep 2024 21:23:03 +0800 Subject: [PATCH 34/64] =?UTF-8?q?0541.=E5=8F=8D=E8=BD=AC=E5=AD=97=E7=AC=A6?= =?UTF-8?q?=E4=B8=B2II=20=E6=80=9D=E8=B7=AF=20=E5=88=A0=E9=99=A4=E4=BA=86?= =?UTF-8?q?=E5=A4=9A=E4=BD=99=E7=9A=84=E2=80=9C=E5=9C=A8=E2=80=9D?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0541.反转字符串II.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md index 3e304fab..5e75d3c3 100644 --- a/problems/0541.反转字符串II.md +++ b/problems/0541.反转字符串II.md @@ -37,7 +37,7 @@ 因为要找的也就是每2 * k 区间的起点，这样写，程序会高效很多。 -**所以当需要固定规律一段一段去处理字符串的时候，要想想在在for循环的表达式上做做文章。** +**所以当需要固定规律一段一段去处理字符串的时候，要想想在for循环的表达式上做做文章。** 性能如下： @@ -505,3 +505,4 @@ impl Solution { + From 2173dd9e6430d8005df0781b4b05a9171208efc4 Mon Sep 17 00:00:00 2001 From: Ziyang Wen Date: Fri, 27 Sep 2024 21:26:55 +0800 Subject: [PATCH 35/64] =?UTF-8?q?0226.Python=E8=BF=AD=E4=BB=A3=E6=B3=95?= =?UTF-8?q?=E5=B9=BF=E5=BA=A6=E4=BC=98=E5=85=88=E9=81=8D=E5=8E=86?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0226.翻转二叉树.md | 10 ++++------ 1 file changed, 4 insertions(+), 6 deletions(-) diff --git a/problems/0226.翻转二叉树.md b/problems/0226.翻转二叉树.md index e501b298..55dc3cbf 100644 --- a/problems/0226.翻转二叉树.md +++ b/problems/0226.翻转二叉树.md @@ -459,11 +459,10 @@ class Solution: queue = collections.deque([root]) while queue: - for i in range(len(queue)): - node = queue.popleft() - node.left, node.right = node.right, node.left - if node.left: queue.append(node.left) - if node.right: queue.append(node.right) + node = queue.popleft() + node.left, node.right = node.right, node.left + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) return root ``` @@ -1033,4 +1032,3 @@ public TreeNode InvertTree(TreeNode root) { - From 9a4dab8f898c175559b121f455707eb82452196f Mon Sep 17 00:00:00 2001 From: Camille <59353274+Camille0512@users.noreply.github.com> Date: Fri, 27 Sep 2024 22:49:01 +0800 Subject: [PATCH 36/64] =?UTF-8?q?Update=200099.=E5=B2=9B=E5=B1=BF=E7=9A=84?= =?UTF-8?q?=E6=95=B0=E9=87=8F=E6=B7=B1=E6=90=9C.md=20Add=20Scala=20code=20?= =?UTF-8?q?to=20solve=20counting=20islands=20problem=20on=20LeetCode?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0099.岛屿的数量深搜.md | 40 +++++++++++++++++++++++ 1 file changed, 40 insertions(+) diff --git a/problems/kamacoder/0099.岛屿的数量深搜.md b/problems/kamacoder/0099.岛屿的数量深搜.md index 6ac7ba3b..da1c0739 100644 --- a/problems/kamacoder/0099.岛屿的数量深搜.md +++ b/problems/kamacoder/0099.岛屿的数量深搜.md @@ -412,6 +412,46 @@ const dfs = (graph, visited, x, y) => { ### Swift ### Scala +```scala +import util.control.Breaks._ + +object Solution { + val dir = List((-1,0), (0,-1), (1,0), (0,1)) // 四个方向 + + def dfs(grid: Array[Array[Char]], visited: Array[Array[Boolean]], row: Int, col: Int): Unit = { + (0 until 4).map { x => + val nextR = row + dir(x)(0) + val nextC = col + dir(x)(1) + breakable { + if(nextR < 0 || nextR >= grid.length || nextC < 0 || nextC >= grid(0).length) break + if (!visited(nextR)(nextC) && grid(nextR)(nextC) == '1') { + visited(nextR)(nextC) = true // 经过就记录 + dfs(grid, visited, nextR, nextC) + } + } + } + } + + def numIslands(grid: Array[Array[Char]]): Int = { + val row = grid.length + val col = grid(0).length + var visited = Array.fill(row)(Array.fill(col)(false)) + var counter = 0 + + (0 until row).map{ r => + (0 until col).map{ c => + if (!visited(r)(c) && grid(r)(c) == '1') { + visited(r)(c) = true // 经过就记录 + dfs(grid, visited, r, c) + counter += 1 + } + } + } + + counter + } +} +``` ### C# From 2f58a7b043eeb19eff88c174366db1777a6fff66 Mon Sep 17 00:00:00 2001 From: DraculaJay <113758447+DraculaJay@users.noreply.github.com> Date: Sat, 28 Sep 2024 04:55:41 +0800 Subject: [PATCH 37/64] =?UTF-8?q?0053=20=E6=B7=BB=E5=8A=A0python=E5=8A=A8?= =?UTF-8?q?=E6=80=81=E8=A7=84=E5=88=92=E8=A7=A3=E6=B3=95=E5=92=8C=E5=8A=A8?= =?UTF-8?q?=E6=80=81=E8=A7=84=E5=88=92=E4=BC=98=E5=8C=96=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0053.最大子序和.md | 36 ++++++++++++++++++++++++++++++++++++ 1 file changed, 36 insertions(+) diff --git a/problems/0053.最大子序和.md b/problems/0053.最大子序和.md index 551c39bf..50757564 100644 --- a/problems/0053.最大子序和.md +++ b/problems/0053.最大子序和.md @@ -229,6 +229,42 @@ class Solution: ``` + +动态规划 + +```python +class Solution: + def maxSubArray(self, nums): + if not nums: + return 0 + dp = [0] * len(nums) # dp[i]表示包括i之前的最大连续子序列和 + dp[0] = nums[0] + result = dp[0] + for i in range(1, len(nums)): + dp[i] = max(dp[i-1]+nums[i], nums[i]) # 状态转移公式 + if dp[i] > result: + result = dp[i] # result 保存dp[i]的最大值 + return result +``` + +动态规划优化 + +```python +class Solution: + def maxSubArray(self, nums: List[int]) -> int: + max_sum = float("-inf") # 初始化结果为负无穷大，方便比较取最大值 + current_sum = 0 # 初始化当前连续和 + + for num in nums: + + # 更新当前连续和 + # 如果原本的连续和加上当前数字之后没有当前数字大，说明原本的连续和是负数，那么就直接从当前数字开始重新计算连续和 + current_sum = max(current_sum+num, num) + max_sum = max(max_sum, current_sum) # 更新结果 + + return max_sum +``` + ### Go 贪心法 ```go From 0ea7587f72abd2c45cc711da696d1c580ddafa06 Mon Sep 17 00:00:00 2001 From: DraculaJay <113758447+DraculaJay@users.noreply.github.com> Date: Sat, 28 Sep 2024 05:04:22 +0800 Subject: [PATCH 38/64] =?UTF-8?q?0055=20=E8=B7=B3=E8=B7=83=E6=B8=B8?= =?UTF-8?q?=E6=88=8F=E6=B7=BB=E5=8A=A0python=20=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0055.跳跃游戏.md | 17 +++++++++++++++++ 1 file changed, 17 insertions(+) diff --git a/problems/0055.跳跃游戏.md b/problems/0055.跳跃游戏.md index 01fd9513..82b433d7 100644 --- a/problems/0055.跳跃游戏.md +++ b/problems/0055.跳跃游戏.md @@ -143,6 +143,23 @@ class Solution: return False ``` +```python +## 基于当前最远可到达位置判断 +class Solution: + def canJump(self, nums: List[int]) -> bool: + far = nums[0] + for i in range(len(nums)): + # 要考虑两个情况 + # 1. i <= far - 表示 当前位置i 可以到达 + # 2. i > far - 表示 当前位置i 无法到达 + if i > far: + return False + far = max(far, nums[i]+i) + # 如果循环正常结束，表示最后一个位置也可以到达，否则会在中途直接退出 + # 关键点在于，要想明白其实列表中的每个位置都是需要验证能否到达的 + return True +``` + ### Go ```go From 754d2d664d848a7f79b7b97347f9fcf27a03749b Mon Sep 17 00:00:00 2001 From: Zhihan Li <54661071+zhihali@users.noreply.github.com> Date: Sun, 29 Sep 2024 23:24:11 +0100 Subject: [PATCH 39/64] =?UTF-8?q?Update=200135.=E5=88=86=E5=8F=91=E7=B3=96?= =?UTF-8?q?=E6=9E=9C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0135.分发糖果.md | 19 +++++++++---------- 1 file changed, 9 insertions(+), 10 deletions(-) diff --git a/problems/0135.分发糖果.md b/problems/0135.分发糖果.md index 6805857e..29eaa06d 100644 --- a/problems/0135.分发糖果.md +++ b/problems/0135.分发糖果.md @@ -177,21 +177,20 @@ class Solution { ```python class Solution: def candy(self, ratings: List[int]) -> int: - candyVec = [1] * len(ratings) + n = len(ratings) + candies = [1] * n - # 从前向后遍历，处理右侧比左侧评分高的情况 - for i in range(1, len(ratings)): + # Forward pass: handle cases where right rating is higher than left + for i in range(1, n): if ratings[i] > ratings[i - 1]: - candyVec[i] = candyVec[i - 1] + 1 + candies[i] = candies[i - 1] + 1 - # 从后向前遍历，处理左侧比右侧评分高的情况 - for i in range(len(ratings) - 2, -1, -1): + # Backward pass: handle cases where left rating is higher than right + for i in range(n - 2, -1, -1): if ratings[i] > ratings[i + 1]: - candyVec[i] = max(candyVec[i], candyVec[i + 1] + 1) + candies[i] = max(candies[i], candies[i + 1] + 1) - # 统计结果 - result = sum(candyVec) - return result + return sum(candies) ``` From 5fb0a12fa14d826ee37179f6807d3e332ae46d76 Mon Sep 17 00:00:00 2001 From: suinming <0223314338aa@gmail.com> Date: Tue, 1 Oct 2024 17:39:24 +0800 Subject: [PATCH 40/64] =?UTF-8?q?feat:=20127.=20=E9=AA=91=E5=A3=AB?= =?UTF-8?q?=E7=9A=84=E6=94=BB=E5=87=BB=E5=A2=9E=E5=8A=A0js=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0126.骑士的攻击astar.md | 125 +++++++++++++++++++++ 1 file changed, 125 insertions(+) diff --git a/problems/kamacoder/0126.骑士的攻击astar.md b/problems/kamacoder/0126.骑士的攻击astar.md index 2d0481ec..1cdba07d 100644 --- a/problems/kamacoder/0126.骑士的攻击astar.md +++ b/problems/kamacoder/0126.骑士的攻击astar.md @@ -375,6 +375,131 @@ for _ in range(n): ### Javascript +```js +class MinHeap { + constructor() { + this.val = [] + } + push(val) { + this.val.push(val) + if (this.val.length > 1) { + this.bubbleUp() + } + } + bubbleUp() { + let pi = this.val.length - 1 + let pp = Math.floor((pi - 1) / 2) + while (pi > 0 && this.val[pp][0] > this.val[pi][0]) { + ;[this.val[pi], this.val[pp]] = [this.val[pp], this.val[pi]] + pi = pp + pp = Math.floor((pi - 1) / 2) + } + } + pop() { + if (this.val.length > 1) { + let pp = 0 + let pi = this.val.length - 1 + ;[this.val[pi], this.val[pp]] = [this.val[pp], this.val[pi]] + const min = this.val.pop() + if (this.val.length > 1) { + this.sinkDown(0) + } + return min + } else if (this.val.length == 1) { + return this.val.pop() + } + + } + sinkDown(parentIdx) { + let pp = parentIdx + let plc = pp * 2 + 1 + let prc = pp * 2 + 2 + let pt = pp // temp pointer + if (plc < this.val.length && this.val[pp][0] > this.val[plc][0]) { + pt = plc + } + if (prc < this.val.length && this.val[pt][0] > this.val[prc][0]) { + pt = prc + } + if (pt != pp) { + ;[this.val[pp], this.val[pt]] = [this.val[pt], this.val[pp]] + this.sinkDown(pt) + } + } +} + +const moves = [ + [1, 2], + [2, 1], + [-1, -2], + [-2, -1], + [-1, 2], + [-2, 1], + [1, -2], + [2, -1] +] + +function dist(a, b) { + return ((a[0] - b[0])**2 + (a[1] - b[1])**2)**0.5 +} + +function isValid(x, y) { + return x >= 1 && y >= 1 && x < 1001 && y < 1001 +} + +function bfs(start, end) { + const step = new Map() + step.set(start.join(" "), 0) + const q = new MinHeap() + q.push([dist(start, end), start[0], start[1]]) + + while(q.val.length) { + const [d, x, y] = q.pop() + // if x and y correspond to end position output result + if (x == end[0] && y == end[1]) { + console.log(step.get(end.join(" "))) + break; + } + for (const [dx, dy] of moves) { + const nx = dx + x + const ny = dy + y + if (isValid(nx, ny)) { + const newStep = step.get([x, y].join(" ")) + 1 + const newDist = dist([nx, ny], [...end]) + const s = step.get([nx, ny].join(" ")) ? + step.get([nx, ny]) : + Number.MAX_VALUE + if (newStep < s) { + q.push( + [ + newStep + newDist, + nx, + ny + ] + ) + step.set([nx, ny].join(" "), newStep) + } + } + } + } +} + +async function main() { + const rl = require('readline').createInterface({ input: process.stdin }) + const iter = rl[Symbol.asyncIterator]() + const readline = async () => (await iter.next()).value + const n = Number((await readline())) + + // find min step + for (let i = 0 ; i < n ; i++) { + const [s1, s2, t1, t2] = (await readline()).split(" ").map(Number) + bfs([s1, s2], [t1, t2]) + } +} + +main() +``` + ### TypeScript ### PhP From 07d219ecd1fb75337d6d5722a1b63635e4f19057 Mon Sep 17 00:00:00 2001 From: Yufan Sheng <18829237653@163.com> Date: Fri, 4 Oct 2024 21:05:14 +1000 Subject: [PATCH 41/64] =?UTF-8?q?=E6=9B=B4=E6=AD=A3=E5=A4=8D=E6=9D=82?= =?UTF-8?q?=E5=BA=A6=E5=88=86=E6=9E=90=E4=B8=AD=E7=9A=84=E7=AC=94=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0452.用最少数量的箭引爆气球.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0452.用最少数量的箭引爆气球.md b/problems/0452.用最少数量的箭引爆气球.md index 318c3035..10fe5771 100644 --- a/problems/0452.用最少数量的箭引爆气球.md +++ b/problems/0452.用最少数量的箭引爆气球.md @@ -110,7 +110,7 @@ public: ``` * 时间复杂度：O(nlog n)，因为有一个快排 -* 空间复杂度：O(1)，有一个快排，最差情况(倒序)时，需要n次递归调用。因此确实需要O(n)的栈空间 +* 空间复杂度：O(n)，有一个快排，最差情况(倒序)时，需要n次递归调用。因此确实需要O(n)的栈空间 可以看出代码并不复杂。 From 2b7fc7d1e183ebcce3930b0235f7f646a6ab3871 Mon Sep 17 00:00:00 2001 From: Yufan Sheng <18829237653@163.com> Date: Fri, 4 Oct 2024 21:22:50 +1000 Subject: [PATCH 42/64] =?UTF-8?q?=E5=88=A0=E9=99=A4=E5=86=97=E4=BD=99?= =?UTF-8?q?=E4=BB=A3=E7=A0=81=EF=BC=8C=E5=A2=9E=E5=8A=A0=E8=AF=A6=E7=BB=86?= =?UTF-8?q?=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0452.用最少数量的箭引爆气球.md | 22 ++++++++++++++-------- 1 file changed, 14 insertions(+), 8 deletions(-) diff --git a/problems/0452.用最少数量的箭引爆气球.md b/problems/0452.用最少数量的箭引爆气球.md index 10fe5771..2099275d 100644 --- a/problems/0452.用最少数量的箭引爆气球.md +++ b/problems/0452.用最少数量的箭引爆气球.md @@ -180,19 +180,25 @@ class Solution: ```python class Solution: # 不改变原数组 def findMinArrowShots(self, points: List[List[int]]) -> int: + if len(points) == 0: + return 0 + points.sort(key = lambda x: x[0]) - sl,sr = points[0][0],points[0][1] + + # points已经按照第一个坐标正序排列，因此只需要设置一个变量，记录右侧坐标（阈值） + # 考虑一个气球范围包含两个不相交气球的情况：气球1: [1, 10], 气球2: [2, 5], 气球3: [6, 10] + curr_min_right = points[0][1] count = 1 + for i in points: - if i[0]>sr: - count+=1 - sl,sr = i[0],i[1] + if i[0] > curr_min_right: + # 当气球左侧大于这个阈值，那么一定就需要在发射一只箭，并且将阈值更新为当前气球的右侧 + count += 1 + curr_min_right = i[1] else: - sl = max(sl,i[0]) - sr = min(sr,i[1]) + # 否则的话，我们只需要求阈值和当前气球的右侧的较小值来更新阈值 + curr_min_right = min(curr_min_right, i[1]) return count - - ``` ### Go ```go From 91137dc297f7a9ca7abb2d59fd9c45afe04f2366 Mon Sep 17 00:00:00 2001 From: Jasonyou <2926593225@qq.com> Date: Thu, 3 Oct 2024 19:39:17 +0800 Subject: [PATCH 43/64] =?UTF-8?q?fix:=20Errata=20=E8=BE=93=E5=87=BA?= =?UTF-8?q?=E8=BE=93=E5=87=BA->=E8=BE=93=E5=85=A5=E8=BE=93=E5=87=BA?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0098.所有可达路径.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/kamacoder/0098.所有可达路径.md b/problems/kamacoder/0098.所有可达路径.md index db35524f..a2f60a90 100644 --- a/problems/kamacoder/0098.所有可达路径.md +++ b/problems/kamacoder/0098.所有可达路径.md @@ -277,7 +277,7 @@ ACM格式大家在输出结果的时候，要关注看看格式问题，特别 有录友可能会想，ACM格式就是麻烦，有空格没有空格有什么影响，结果对了不就行了？ -ACM模式相对于核心代码模式（力扣） 更考验大家对代码的掌控能力。 例如工程代码里，输出输出都是要自己控制的。这也是为什么大公司笔试，都是ACM模式。 +ACM模式相对于核心代码模式（力扣） 更考验大家对代码的掌控能力。 例如工程代码里，输入输出都是要自己控制的。这也是为什么大公司笔试，都是ACM模式。 以上代码中，结果都存在了 result数组里（二维数组，每一行是一个结果），最后将其打印出来。（重点看注释） From 4eef0863a293b7f2cfd36214a3281ae26093022e Mon Sep 17 00:00:00 2001 From: Jasonyou <2926593225@qq.com> Date: Fri, 4 Oct 2024 21:18:01 +0800 Subject: [PATCH 44/64] =?UTF-8?q?feat:=200100.=E5=B2=9B=E5=B1=BF=E6=9C=80?= =?UTF-8?q?=E5=A4=A7=E9=9D=A2=E7=A7=AF=20Java=E5=AE=9E=E7=8E=B0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0100.岛屿的最大面积.md | 123 +++++++++++++++++++++- 1 file changed, 122 insertions(+), 1 deletion(-) diff --git a/problems/kamacoder/0100.岛屿的最大面积.md b/problems/kamacoder/0100.岛屿的最大面积.md index 51bfc57f..871925df 100644 --- a/problems/kamacoder/0100.岛屿的最大面积.md +++ b/problems/kamacoder/0100.岛屿的最大面积.md @@ -222,7 +222,128 @@ public: ## 其他语言版本 -### Java +### Java + +```java +import java.util.*; +import java.math.*; + +/** + * DFS版 + */ +public class Main{ + + static final int[][] dir={{0,1},{1,0},{0,-1},{-1,0}}; + static int result=0; + static int count=0; + + public static void main(String[] args){ + Scanner scanner = new Scanner(System.in); + int n = scanner.nextInt(); + int m = scanner.nextInt(); + int[][] map = new int[n][m]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < m; j++) { + map[i][j]=scanner.nextInt(); + } + } + boolean[][] visited = new boolean[n][m]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < m; j++) { + if(!visited[i][j]&&map[i][j]==1){ + count=0; + dfs(map,visited,i,j); + result= Math.max(count, result); + } + } + } + System.out.println(result); + } + + static void dfs(int[][] map,boolean[][] visited,int x,int y){ + count++; + visited[x][y]=true; + for (int i = 0; i < 4; i++) { + int nextX=x+dir[i][0]; + int nextY=y+dir[i][1]; + //水或者已经访问过的跳过 + if(nextX<0||nextY<0 + ||nextX>=map.length||nextY>=map[0].length + ||visited[nextX][nextY]||map[nextX][nextY]==0)continue; + + dfs(map,visited,nextX,nextY); + } + } +} +``` + +```java +import java.util.*; +import java.math.*; + +/** + * BFS版 + */ +public class Main { + static class Node { + int x; + int y; + + public Node(int x, int y) { + this.x = x; + this.y = y; + } + } + + static final int[][] dir = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; + static int result = 0; + static int count = 0; + + public static void main(String[] args) { + Scanner scanner = new Scanner(System.in); + int n = scanner.nextInt(); + int m = scanner.nextInt(); + int[][] map = new int[n][m]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < m; j++) { + map[i][j] = scanner.nextInt(); + } + } + boolean[][] visited = new boolean[n][m]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < m; j++) { + if (!visited[i][j] && map[i][j] == 1) { + count = 0; + bfs(map, visited, i, j); + result = Math.max(count, result); + } + } + } + System.out.println(result); + } + + static void bfs(int[][] map, boolean[][] visited, int x, int y) { + Queue q = new LinkedList<>(); + q.add(new Node(x, y)); + visited[x][y] = true; + count++; + while (!q.isEmpty()) { + Node node = q.remove(); + for (int i = 0; i < 4; i++) { + int nextX = node.x + dir[i][0]; + int nextY = node.y + dir[i][1]; + if (nextX < 0 || nextY < 0 || nextX >= map.length || nextY >= map[0].length || visited[nextX][nextY] || map[nextX][nextY] == 0) + continue; + q.add(new Node(nextX, nextY)); + visited[nextX][nextY] = true; + count++; + } + } + } +} + +``` + ### Python From fc004742e70956263fbfba9ca48c3b80d6aee845 Mon Sep 17 00:00:00 2001 From: "Ethan.Liu" Date: Sat, 5 Oct 2024 12:26:05 +0800 Subject: [PATCH 45/64] =?UTF-8?q?feat:=200977.=E6=9C=89=E5=BA=8F=E6=95=B0?= =?UTF-8?q?=E7=BB=84=E7=9A=84=E5=B9=B3=E6=96=B9=20python3=E5=8F=8C?= =?UTF-8?q?=E6=8C=87=E9=92=88=E4=B8=89=E6=AD=A5=E4=BC=98=E5=8C=96?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0977.有序数组的平方.md | 38 +++++++++++++++++++++++++++++++++ 1 file changed, 38 insertions(+) diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md index effa9055..1a6604c2 100644 --- a/problems/0977.有序数组的平方.md +++ b/problems/0977.有序数组的平方.md @@ -100,6 +100,7 @@ public: ## 其他语言版本 ### Java： + 排序法 ```Java class Solution { @@ -209,6 +210,43 @@ class Solution: return new_list[::-1] ``` +```python3 +(双指针优化版本） 三步优化 + class Solution: + def sortedSquares(self, nums: List[int]) -> List[int]: + """ + 整体思想：有序数组的绝对值最大值永远在两头，比较两头，平方大的插到新数组的最后 + 优 化：1. 优化所有元素为非正或非负的情况 + 2. 头尾平方的大小比较直接将头尾相加与0进行比较即可 + 3. 新的平方排序数组的插入索引可以用倒序插入实现（针对for循环，while循环不适用） + """ + + # 特殊情况, 元素都非负（优化1） + if nums[0] >= 0: + return [num ** 2 for num in nums] # 按顺序平方即可 + # 最后一个非正，全负有序的 + if nums[-1] <= 0: + return [x ** 2 for x in nums[::-1]] # 倒序平方后的数组 + + # 一般情况, 有正有负 + i = 0 # 原数组头索引 + j = len(nums) - 1 # 原数组尾部索引 + new_nums = [0] * len(nums) # 新建一个等长数组用于保存排序后的结果 + # end_index = len(nums) - 1 # 新的排序数组(是新数组)尾插索引, 每次需要减一（优化3优化了） + + for end_index in range(len(nums)-1, -1, -1): # (优化3，倒序，不用单独创建变量) + # if nums[i] ** 2 >= nums[j] ** 2: + if nums[i] + nums[j] <= 0: # (优化2) + new_nums[end_index] = nums[i] ** 2 + i += 1 + # end_index -= 1 (优化3) + else: + new_nums[end_index] = nums[j] ** 2 + j -= 1 + # end_index -= 1 (优化3) + return new_nums +``` + ### Go： ```Go From 0a33a8d05c1c430bdb56d63a360e2de0d09b6ace Mon Sep 17 00:00:00 2001 From: Nomop <13098939400@163.com> Date: Sat, 5 Oct 2024 15:21:32 +0800 Subject: [PATCH 46/64] =?UTF-8?q?Update=200100.=E5=B2=9B=E5=B1=BF=E7=9A=84?= =?UTF-8?q?=E6=9C=80=E5=A4=A7=E9=9D=A2=E7=A7=AF.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 100.岛屿的最大面积新增Rust解法 --- problems/kamacoder/0100.岛屿的最大面积.md | 138 ++++++++++++++++++++++ 1 file changed, 138 insertions(+) diff --git a/problems/kamacoder/0100.岛屿的最大面积.md b/problems/kamacoder/0100.岛屿的最大面积.md index 51bfc57f..50e40f3a 100644 --- a/problems/kamacoder/0100.岛屿的最大面积.md +++ b/problems/kamacoder/0100.岛屿的最大面积.md @@ -389,6 +389,144 @@ func main() { ### Rust +DFS + +``` rust +use std::io; +use std::cmp; + +// 定义四个方向 +const DIRECTIONS: [(i32, i32); 4] = [(0, 1), (1, 0), (-1, 0), (0, -1)]; + +fn dfs(grid: &Vec>, visited: &mut Vec>, x: usize, y: usize, count: &mut i32) { + if visited[x][y] || grid[x][y] == 0 { + return; // 终止条件：已访问或者遇到海水 + } + visited[x][y] = true; // 标记已访问 + *count += 1; + + for &(dx, dy) in DIRECTIONS.iter() { + let new_x = x as i32 + dx; + let new_y = y as i32 + dy; + + // 检查边界条件 + if new_x >= 0 && new_x < grid.len() as i32 && new_y >= 0 && new_y < grid[0].len() as i32 { + dfs(grid, visited, new_x as usize, new_y as usize, count); + } + } +} + +fn main() { + let mut input = String::new(); + + // 读取 n 和 m + io::stdin().read_line(&mut input); + let dims: Vec = input.trim().split_whitespace().map(|s| s.parse().unwrap()).collect(); + let (n, m) = (dims[0], dims[1]); + + // 读取 grid + let mut grid = vec![]; + for _ in 0..n { + input.clear(); + io::stdin().read_line(&mut input); + let row: Vec = input.trim().split_whitespace().map(|s| s.parse().unwrap()).collect(); + grid.push(row); + } + + // 初始化访问记录 + let mut visited = vec![vec![false; m]; n]; + let mut result = 0; + + // 遍历所有格子 + for i in 0..n { + for j in 0..m { + if !visited[i][j] && grid[i][j] == 1 { + let mut count = 0; + dfs(&grid, &mut visited, i, j, &mut count); + result = cmp::max(result, count); + } + } + } + + // 输出结果 + println!("{}", result); +} + +``` +BFS +```rust +use std::io; +use std::collections::VecDeque; + +// 定义四个方向 +const DIRECTIONS: [(i32, i32); 4] = [(0, 1), (1, 0), (-1, 0), (0, -1)]; + +fn bfs(grid: &Vec>, visited: &mut Vec>, x: usize, y: usize) -> i32 { + let mut count = 0; + let mut queue = VecDeque::new(); + queue.push_back((x, y)); + visited[x][y] = true; // 标记已访问 + + while let Some((cur_x, cur_y)) = queue.pop_front() { + count += 1; // 增加计数 + + for &(dx, dy) in DIRECTIONS.iter() { + let new_x = cur_x as i32 + dx; + let new_y = cur_y as i32 + dy; + + // 检查边界条件 + if new_x >= 0 && new_x < grid.len() as i32 && new_y >= 0 && new_y < grid[0].len() as i32 { + let new_x_usize = new_x as usize; + let new_y_usize = new_y as usize; + + // 如果未访问且是陆地，加入队列 + if !visited[new_x_usize][new_y_usize] && grid[new_x_usize][new_y_usize] == 1 { + visited[new_x_usize][new_y_usize] = true; // 标记已访问 + queue.push_back((new_x_usize, new_y_usize)); + } + } + } + } + + count +} + +fn main() { + let mut input = String::new(); + + // 读取 n 和 m + io::stdin().read_line(&mut input).expect("Failed to read line"); + let dims: Vec = input.trim().split_whitespace().map(|s| s.parse().unwrap()).collect(); + let (n, m) = (dims[0], dims[1]); + + // 读取 grid + let mut grid = vec![]; + for _ in 0..n { + input.clear(); + io::stdin().read_line(&mut input).expect("Failed to read line"); + let row: Vec = input.trim().split_whitespace().map(|s| s.parse().unwrap()).collect(); + grid.push(row); + } + + // 初始化访问记录 + let mut visited = vec![vec![false; m]; n]; + let mut result = 0; + + // 遍历所有格子 + for i in 0..n { + for j in 0..m { + if !visited[i][j] && grid[i][j] == 1 { + let count = bfs(&grid, &mut visited, i, j); + result = result.max(count); + } + } + } + + // 输出结果 + println!("{}", result); +} + +``` ### Javascript From a663089383732a56011c1b630c30635265c316a7 Mon Sep 17 00:00:00 2001 From: jjblaack Date: Sat, 5 Oct 2024 18:07:00 +0800 Subject: [PATCH 47/64] =?UTF-8?q?=E5=B2=9B=E5=B1=BF=E7=9A=84=E6=9C=80?= =?UTF-8?q?=E5=A4=A7=E9=9D=A2=E7=A7=AFjava=E5=AE=9E=E7=8E=B0=EF=BC=88dfs+b?= =?UTF-8?q?fs=EF=BC=89?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0100.岛屿的最大面积.md | 116 +++++++++++++++++++++- 1 file changed, 115 insertions(+), 1 deletion(-) diff --git a/problems/kamacoder/0100.岛屿的最大面积.md b/problems/kamacoder/0100.岛屿的最大面积.md index 51bfc57f..38b98d0e 100644 --- a/problems/kamacoder/0100.岛屿的最大面积.md +++ b/problems/kamacoder/0100.岛屿的最大面积.md @@ -223,7 +223,121 @@ public: ## 其他语言版本 ### Java - +DFS +```java +//这里的实现为主函数处理每个岛屿的第一块陆地 方式 +//所以是主函数直接置count为1，剩余的交给dfs来做。 +import java.util.*; +public class Main{ + static int[][] dir = {{0,-1}, {1,0}, {0,1}, {-1, 0}};//四个方向 + static int count = 0; + public static void dfs(boolean[][] visited, int x, int y, int[][] grid){ + for(int i = 0; i < 4; i++){ + int nextX = x + dir[i][0]; + int nextY = y + dir[i][1]; + if(nextX < 0 || nextY < 0 || nextY >= grid[0].length || nextX >= grid.length){ + continue; + } + if(!visited[nextX][nextY] && grid[nextX][nextY] == 1){ + count++; + visited[nextX][nextY] = true; + dfs(visited, nextX, nextY, grid); + } + } + } + public static void main(String[] args){ + Scanner in = new Scanner(System.in); + int n = in.nextInt(); + int m = in.nextInt(); + int[][] grid = new int[n][m]; + for(int i = 0; i < n; i++){ + for(int j = 0; j < m; j++){ + grid[i][j] = in.nextInt(); + } + } + + int result = 0; + boolean[][] visited = new boolean[n][m]; + for(int i = 0; i < n; i++){ + for(int j = 0; j < m; j++){ + if(!visited[i][j] && grid[i][j] == 1){ + visited[i][j] = true; + count = 1; + dfs(visited, i, j, grid); + //dfs遍历完了一座岛屿，就比较count和result，保留最大的 + result = Math.max(result, count); + } + } + } + System.out.println(result); + } +} +``` +BFS +```java +import java.util.*; +public class Main{ + static int[][] dir = {{0,-1}, {1,0}, {0,1}, {-1, 0}};//下右上左的顺序 + static int count = 0; + public static void bfs(boolean[][] visited, int x, int y, int[][] grid){ + Queue queue = new LinkedList(); + queue.add(new pair(x,y)); + count = 1; //该岛屿的第一块陆地被visit了 + + //对这个岛屿的所有都入队，除非上下左右都没有未访问的陆地 + while(!queue.isEmpty()){ + int curX = queue.peek().x; + int curY = queue.poll().y; + //对每块陆地都进行上下左右的入队和计算（遍历），自然就是按广度优先了 + for(int i = 0; i < 4; i++){ + int nextX = curX + dir[i][0]; + int nextY = curY + dir[i][1]; + if(nextX < 0 || nextY < 0 || nextX >= grid.length || nextY >= grid[0].length){ + continue; + } + if(!visited[nextX][nextY] && grid[nextX][nextY] == 1){ + count++; + queue.add(new pair(nextX, nextY)); + visited[nextX][nextY] = true; + } + } + } + } + + static class pair{ + int x; + int y; + pair(int x, int y){ + this.x = x; + this.y = y; + } + } + + public static void main(String[] args){ + Scanner in = new Scanner(System.in); + int n = in.nextInt(); + int m = in.nextInt(); + int[][] grid = new int[n][m]; + for(int i = 0; i < n; i++){ + for(int j = 0; j < m; j++){ + grid[i][j] = in.nextInt(); + } + } + int result = 0; + boolean[][] visited = new boolean[n][m]; + for(int i = 0; i < n; i++){ + for(int j = 0; j < m; j++){ + if(!visited[i][j] && grid[i][j] == 1){ + visited[i][j] = true; + bfs(visited, i, j, grid); + result = Math.max(result, count); + } + } + } + System.out.println(result); + } +} +``` ### Python DFS From 9c7131e253ad9b59aaa69f102f03e0154752b660 Mon Sep 17 00:00:00 2001 From: Jasonyou <2926593225@qq.com> Date: Sun, 6 Oct 2024 03:04:09 +0800 Subject: [PATCH 48/64] =?UTF-8?q?feate:=20=E5=AD=97=E7=AC=A6=E4=B8=B2?= =?UTF-8?q?=E6=8E=A5=E9=BE=99=20=20Java=E4=BC=98=E5=8C=96=E7=89=88?= =?UTF-8?q?=E5=B9=B6=E4=BF=AE=E5=A4=8D=E5=AD=98=E5=9C=A8=E9=97=AE=E9=A2=98?= =?UTF-8?q?=201.=E4=BD=BF=E7=94=A8null=E6=A0=87=E5=BF=97=E5=88=86=E5=B1=82?= =?UTF-8?q?=202.=E5=AF=BB=E6=89=BE=E9=82=BB=E6=8E=A5=E8=8A=82=E7=82=B9?= =?UTF-8?q?=E6=97=B6=EF=BC=8C=E5=9B=9E=E6=BB=9A=E4=BF=AE=E6=94=B9=EF=BC=8C?= =?UTF-8?q?=E9=81=BF=E5=85=8D=E5=A4=9A=E6=AC=A1=E6=95=B0=E7=BB=84=E5=A4=8D?= =?UTF-8?q?=E5=88=B6=203.=E4=BF=AE=E5=A4=8D=E8=BE=B9=E7=95=8C=E6=9D=A1?= =?UTF-8?q?=E4=BB=B6=EF=BC=9AbeginStr=3D=3DendStr=E6=97=B6=EF=BC=8C?= =?UTF-8?q?=E5=BA=94=E8=AF=A5=E8=BE=93=E5=87=BA0=EF=BC=9B?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0110.字符串接龙.md | 106 +++++++++++++------------- 1 file changed, 55 insertions(+), 51 deletions(-) diff --git a/problems/kamacoder/0110.字符串接龙.md b/problems/kamacoder/0110.字符串接龙.md index 308e50b7..ef261c4d 100644 --- a/problems/kamacoder/0110.字符串接龙.md +++ b/problems/kamacoder/0110.字符串接龙.md @@ -152,66 +152,70 @@ int main() { ## 其他语言版本 -### Java +### Java + ```Java +import java.util.*; + public class Main { - // BFS方法 - public static int ladderLength(String beginWord, String endWord, List wordList) { - // 使用set作为查询容器，效率更高 - HashSet set = new HashSet<>(wordList); - - // 声明一个queue存储每次变更一个字符得到的且存在于容器中的新字符串 - Queue queue = new LinkedList<>(); - - // 声明一个hashMap存储遍历到的字符串以及所走过的路径path - HashMap visitMap = new HashMap<>(); - queue.offer(beginWord); - visitMap.put(beginWord, 1); - - while (!queue.isEmpty()) { - String curWord = queue.poll(); - int path = visitMap.get(curWord); + public static void main(String[] args) { + Scanner scanner = new Scanner(System.in); + int n = scanner.nextInt(); + scanner.nextLine(); + String beginStr = scanner.next(); + String endStr = scanner.next(); + scanner.nextLine(); + List wordList = new ArrayList<>(); + wordList.add(beginStr); + wordList.add(endStr); + for (int i = 0; i < n; i++) { + wordList.add(scanner.nextLine()); + } + int count = bfs(beginStr, endStr, wordList); + System.out.println(count); + } - for (int i = 0; i < curWord.length(); i++) { - char[] ch = curWord.toCharArray(); - // 每个位置尝试26个字母 - for (char k = 'a'; k <= 'z'; k++) { - ch[i] = k; - - String newWord = new String(ch); - if (newWord.equals(endWord)) return path + 1; - - // 如果这个新字符串存在于容器且之前未被访问到 - if (set.contains(newWord) && !visitMap.containsKey(newWord)) { - visitMap.put(newWord, path + 1); - queue.offer(newWord); + /** + * 广度优先搜索-寻找最短路径 + */ + public static int bfs(String beginStr, String endStr, List wordList) { + int len = 1; + Set set = new HashSet<>(wordList); + Set visited = new HashSet<>(); + Queue q = new LinkedList<>(); + visited.add(beginStr); + q.add(beginStr); + q.add(null); + while (!q.isEmpty()) { + String node = q.remove(); + //上一层结束，若下一层还有节点进入下一层 + if (node == null) { + if (!q.isEmpty()) { + len++; + q.add(null); + } + continue; + } + char[] charArray = node.toCharArray(); + //寻找邻接节点 + for (int i = 0; i < charArray.length; i++) { + //记录旧值，用于回滚修改 + char old = charArray[i]; + for (char j = 'a'; j <= 'z'; j++) { + charArray[i] = j; + String newWord = new String(charArray); + if (set.contains(newWord) && !visited.contains(newWord)) { + q.add(newWord); + visited.add(newWord); + //找到结尾 + if (newWord.equals(endStr)) return len + 1; } } + charArray[i] = old; } } - return 0; } - - public static void main (String[] args) { - /* code */ - // 接收输入 - Scanner sc = new Scanner(System.in); - int N = sc.nextInt(); - sc.nextLine(); - String[] strs = sc.nextLine().split(" "); - - List wordList = new ArrayList<>(); - for (int i = 0; i < N; i++) { - wordList.add(sc.nextLine()); - } - - // wordList.add(strs[1]); - - // 打印结果 - int result = ladderLength(strs[0], strs[1], wordList); - System.out.println(result); - } } ``` From d3804674e4b26d29800720a35d42c177a5a49e66 Mon Sep 17 00:00:00 2001 From: Jasonyou <2926593225@qq.com> Date: Mon, 7 Oct 2024 06:29:20 +0800 Subject: [PATCH 49/64] =?UTF-8?q?perf=EF=BC=9A=E5=B0=81=E8=A3=85=E6=88=90?= =?UTF-8?q?=E7=B1=BB=EF=BC=8C=E7=90=86=E8=A7=A3=E8=B5=B7=E6=9D=A5=E6=9B=B4?= =?UTF-8?q?=E5=AE=B9=E6=98=93?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0109.冗余连接II.md | 182 +++++++++++++++----------- 1 file changed, 104 insertions(+), 78 deletions(-) diff --git a/problems/kamacoder/0109.冗余连接II.md b/problems/kamacoder/0109.冗余连接II.md index 2bd4eac6..56959d87 100644 --- a/problems/kamacoder/0109.冗余连接II.md +++ b/problems/kamacoder/0109.冗余连接II.md @@ -250,105 +250,131 @@ int main() { ## 其他语言版本 ### Java + ```java -import java.util.ArrayList; -import java.util.List; -import java.util.Scanner; +import java.util.*; +/* + * 冗余连接II。主要问题是存在入度为2或者成环，也可能两个问题同时存在。 + * 1.判断入度为2的边 + * 2.判断是否成环（并查集） + */ + public class Main { - static int n; - static int[] father = new int[1001]; // 并查集数组 + /** + * 并查集模板 + */ + static class Disjoint { - // 并查集初始化 - public static void init() { - for (int i = 1; i <= n; ++i) { - father[i] = i; - } - } + private final int[] father; - // 并查集里寻根的过程 - public static int find(int u) { - if (u == father[u]) return u; - return father[u] = find(father[u]); // 路径压缩 - } - - // 将 v->u 这条边加入并查集 - public static void join(int u, int v) { - u = find(u); - v = find(v); - if (u != v) { - father[v] = u; // 合并两棵树 - } - } - - // 判断 u 和 v 是否有同一个根 - public static boolean same(int u, int v) { - return find(u) == find(v); - } - - // 在有向图里找到删除的那条边，使其变成树 - public static void getRemoveEdge(List edges) { - init(); // 初始化并查集 - for (int i = 0; i < n; i++) { // 遍历所有的边 - if (same(edges.get(i)[0], edges.get(i)[1])) { // 如果构成有向环了，就是要删除的边 - System.out.println(edges.get(i)[0] + " " + edges.get(i)[1]); - return; - } else { - join(edges.get(i)[0], edges.get(i)[1]); + public Disjoint(int n) { + father = new int[n]; + for (int i = 0; i < n; i++) { + father[i] = i; } } + + public void join(int n, int m) { + n = find(n); + m = find(m); + if (n == m) return; + father[n] = m; + } + + public int find(int n) { + return father[n] == n ? n : (father[n] = find(father[n])); + } + + public boolean isSame(int n, int m) { + return find(n) == find(m); + } } - // 删一条边之后判断是不是树 - public static boolean isTreeAfterRemoveEdge(List edges, int deleteEdge) { - init(); // 初始化并查集 + static class Edge { + int s; + int t; + + public Edge(int s, int t) { + this.s = s; + this.t = t; + } + } + + static class Node { + int id; + int in; + int out; + } + + public static void main(String[] args) { + Scanner scanner = new Scanner(System.in); + int n = scanner.nextInt(); + List edges = new ArrayList<>(); + Node[] nodeMap = new Node[n + 1]; + for (int i = 1; i <= n; i++) { + nodeMap[i] = new Node(); + } + Integer doubleIn = null; for (int i = 0; i < n; i++) { - if (i == deleteEdge) continue; - if (same(edges.get(i)[0], edges.get(i)[1])) { // 如果构成有向环了，一定不是树 + int s = scanner.nextInt(); + int t = scanner.nextInt(); + //记录入度 + nodeMap[t].in++; + if (!(nodeMap[t].in < 2)) doubleIn = t; + Edge edge = new Edge(s, t); + edges.add(edge); + } + Edge result = null; + //存在入度为2的节点，既要消除入度为2的问题同时解除可能存在的环 + if (doubleIn != null) { + List doubleInEdges = new ArrayList<>(); + for (Edge edge : edges) { + if (edge.t == doubleIn) doubleInEdges.add(edge); + if (doubleInEdges.size() == 2) break; + } + Edge edge = doubleInEdges.get(1); + if (isTreeWithExclude(edges, edge, nodeMap)) { + result = edge; + } else { + result = doubleInEdges.get(0); + } + } else { + //不存在入度为2的节点,则只需要解除环即可 + result = getRemoveEdge(edges, nodeMap); + } + + System.out.println(result.s + " " + result.t); + } + + public static boolean isTreeWithExclude(List edges, Edge exculdEdge, Node[] nodeMap) { + Disjoint disjoint = new Disjoint(nodeMap.length + 1); + for (Edge edge : edges) { + if (edge == exculdEdge) continue; + //成环则不是树 + if (disjoint.isSame(edge.s, edge.t)) { return false; } - join(edges.get(i)[0], edges.get(i)[1]); + disjoint.join(edge.s, edge.t); } return true; } - public static void main(String[] args) { - Scanner sc = new Scanner(System.in); - List edges = new ArrayList<>(); // 存储所有的边 + public static Edge getRemoveEdge(List edges, Node[] nodeMap) { + int length = nodeMap.length; + Disjoint disjoint = new Disjoint(length); - n = sc.nextInt(); // 顶点数 - int[] inDegree = new int[n + 1]; // 记录每个节点的入度 - for (int i = 0; i < n; i++) { - int s = sc.nextInt(); // 边的起点 - int t = sc.nextInt(); // 边的终点 - inDegree[t]++; - edges.add(new int[]{s, t}); // 将边加入列表 + for (Edge edge : edges) { + if (disjoint.isSame(edge.s, edge.t)) return edge; + disjoint.join(edge.s, edge.t); } - - List vec = new ArrayList<>(); // 记录入度为2的边（如果有的话就两条边） - // 找入度为2的节点所对应的边，注意要倒序，因为优先删除最后出现的一条边 - for (int i = n - 1; i >= 0; i--) { - if (inDegree[edges.get(i)[1]] == 2) { - vec.add(i); - } - } - - // 情况一、情况二 - if (vec.size() > 0) { - // vec里的边已经按照倒叙放的，所以优先删 vec.get(0) 这条边 - if (isTreeAfterRemoveEdge(edges, vec.get(0))) { - System.out.println(edges.get(vec.get(0))[0] + " " + edges.get(vec.get(0))[1]); - } else { - System.out.println(edges.get(vec.get(1))[0] + " " + edges.get(vec.get(1))[1]); - } - return; - } - - // 处理情况三：明确没有入度为2的情况，一定有有向环，找到构成环的边返回即可 - getRemoveEdge(edges); + return null; } + } + ``` + ### Python ```python From 8669f2ad20f3c7f550557584fa57b4622e5862e7 Mon Sep 17 00:00:00 2001 From: Leehouc <152672308+Leehouc@users.noreply.github.com> Date: Mon, 7 Oct 2024 13:19:27 +0800 Subject: [PATCH 50/64] =?UTF-8?q?=E4=BF=AE=E6=94=B90459.=E9=87=8D=E5=A4=8D?= =?UTF-8?q?=E7=9A=84=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md=E9=94=99?= =?UTF-8?q?=E5=88=AB=E5=AD=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 打码->代码 --- problems/0459.重复的子字符串.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index dbf32e64..5c86e122 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -312,7 +312,7 @@ next 数组记录的就是最长相同前后缀（ [字符串：KMP算法精讲] 4可以被 12(字符串的长度) 整除，所以说明有重复的子字符串（asdf）。 -### 打码实现 +### 代码实现 C++代码如下：（这里使用了前缀表统一减一的实现方式） From fdae38cf7e575cc01f3db7807ca48b942dcf2cd2 Mon Sep 17 00:00:00 2001 From: markwang Date: Tue, 8 Oct 2024 15:16:23 +0800 Subject: [PATCH 51/64] =?UTF-8?q?337.=E6=89=93=E5=AE=B6=E5=8A=AB=E8=88=8DI?= =?UTF-8?q?II=E5=A2=9E=E5=8A=A0Go=E9=80=92=E5=BD=92=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0337.打家劫舍III.md | 84 ++++++++++++++++++++++++++++++++++++ 1 file changed, 84 insertions(+) diff --git a/problems/0337.打家劫舍III.md b/problems/0337.打家劫舍III.md index 7aae5cbf..12e31aba 100644 --- a/problems/0337.打家劫舍III.md +++ b/problems/0337.打家劫舍III.md @@ -388,6 +388,90 @@ class Solution: ### Go +暴力递归 + +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +func rob(root *TreeNode) int { + if root == nil { + return 0 + } + if root.Left == nil && root.Right == nil { + return root.Val + } + // 偷父节点 + val1 := root.Val + if root.Left != nil { + val1 += rob(root.Left.Left) + rob(root.Left.Right) // 跳过root->left，相当于不考虑左孩子了 + } + if root.Right != nil { + val1 += rob(root.Right.Left) + rob(root.Right.Right) // 跳过root->right，相当于不考虑右孩子了 + } + // 不偷父节点 + val2 := rob(root.Left) + rob(root.Right) // 考虑root的左右孩子 + return max(val1, val2) +} + +func max(x, y int) int { + if x > y { + return x + } + return y +} +``` + +记忆化递推 + +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +var umap = make(map[*TreeNode]int) + +func rob(root *TreeNode) int { + if root == nil { + return 0 + } + if root.Left == nil && root.Right == nil { + return root.Val + } + if val, ok := umap[root]; ok { + return val // 如果umap里已经有记录则直接返回 + } + // 偷父节点 + val1 := root.Val + if root.Left != nil { + val1 += rob(root.Left.Left) + rob(root.Left.Right) // 跳过root->left，相当于不考虑左孩子了 + } + if root.Right != nil { + val1 += rob(root.Right.Left) + rob(root.Right.Right) // 跳过root->right，相当于不考虑右孩子了 + } + // 不偷父节点 + val2 := rob(root.Left) + rob(root.Right) // 考虑root的左右孩子 + umap[root] = max(val1, val2) // umap记录一下结果 + return max(val1, val2) +} + +func max(x, y int) int { + if x > y { + return x + } + return y +} +``` + 动态规划 ```go From d33c76e74aa90b8ba319af04b0c7f731f6bfabca Mon Sep 17 00:00:00 2001 From: 01geek Date: Wed, 9 Oct 2024 16:28:31 +0800 Subject: [PATCH 52/64] =?UTF-8?q?=E7=BA=A0=E6=AD=A3=20problems/1047.?= =?UTF-8?q?=E5=88=A0=E9=99=A4=E5=AD=97=E7=AC=A6=E4=B8=B2=E4=B8=AD=E7=9A=84?= =?UTF-8?q?=E6=89=80=E6=9C=89=E7=9B=B8=E9=82=BB=E9=87=8D=E5=A4=8D=E9=A1=B9?= =?UTF-8?q?.md=20=E4=B8=ADjava=E7=89=88=E6=9C=AC=E7=9B=B8=E5=85=B3?= =?UTF-8?q?=E6=B3=A8=E9=87=8A=EF=BC=9A=E5=B0=86top>0=E7=BA=A0=E6=AD=A3?= =?UTF-8?q?=E4=B8=BAtop>=3D0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1047.删除字符串中的所有相邻重复项.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/1047.删除字符串中的所有相邻重复项.md b/problems/1047.删除字符串中的所有相邻重复项.md index 7232008a..51ec4e62 100644 --- a/problems/1047.删除字符串中的所有相邻重复项.md +++ b/problems/1047.删除字符串中的所有相邻重复项.md @@ -164,7 +164,7 @@ class Solution { int top = -1; for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); - // 当 top > 0,即栈中有字符时，当前字符如果和栈中字符相等，弹出栈顶字符，同时 top-- + // 当 top >= 0,即栈中有字符时，当前字符如果和栈中字符相等，弹出栈顶字符，同时 top-- if (top >= 0 && res.charAt(top) == c) { res.deleteCharAt(top); top--; From ba2062471b0a041b748bb543d58514ceb849b21f Mon Sep 17 00:00:00 2001 From: programmercarl <826123027@qq.com> Date: Wed, 9 Oct 2024 16:58:09 +0800 Subject: [PATCH 53/64] Update --- problems/0459.重复的子字符串.md | 22 +++++++++++----------- problems/0518.零钱兑换II.md | 28 ++++++++++++++++++++++++---- 2 files changed, 35 insertions(+), 15 deletions(-) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index 254d921d..bee1c102 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -64,7 +64,7 @@ 如果有一个字符串s，在 s + s 拼接后， 不算首尾字符，如果能凑成s字符串，说明s 一定是重复子串组成。 -如图，字符串s，图中数字为数组下标，在 s + s 拼接后， 不算首尾字符，中间凑成s字符串。 +如图，字符串s，图中数字为数组下标，在 s + s 拼接后， 不算首尾字符，中间凑成s字符串。 （图中数字为数组下标） ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20240910115555.png) @@ -163,9 +163,7 @@ KMP算法中next数组为什么遇到字符不匹配的时候可以找到上一 如果一个字符串s是由重复子串组成，那么 最长相等前后缀不包含的子串一定是字符串s的最小重复子串。 -证明： 如果s 是有是有最小重复子串p组成。 - -即 s = n * p +如果s 是由最小重复子串p组成，即 s = n * p 那么相同前后缀可以是这样： @@ -203,12 +201,14 @@ p2 = p1，p3 = p2 即： p1 = p2 = p3 最长相等前后缀不包含的子串已经是字符串s的最小重复子串，那么字符串s一定由重复子串组成，这个不需要证明了。 -关键是要要证明：最长相等前后缀不包含的子串什么时候才是字符串s的最小重复子串呢。 +关键是要证明：最长相等前后缀不包含的子串什么时候才是字符串s的最小重复子串呢。 -情况一， 最长相等前后缀不包含的子串的长度 比 字符串s的一半的长度还大，那一定不是字符串s的重复子串 +情况一， 最长相等前后缀不包含的子串的长度 比 字符串s的一半的长度还大，那一定不是字符串s的重复子串，如图： ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20240911110236.png) +图中：前后缀不包含的子串的长度 大于 字符串s的长度的 二分之一 + -------------- 情况二，最长相等前后缀不包含的子串的长度 可以被 字符串s的长度整除，如图： @@ -230,7 +230,7 @@ p2 = p1，p3 = p2 即： p1 = p2 = p3 即 s[0]s[1] 是最小重复子串 -以上推导中，录友可能想，你怎么知道 s[0] 和 s[1] 就不相同呢？ s[0] 为什么就不能使最小重复子串。 +以上推导中，录友可能想，你怎么知道 s[0] 和 s[1] 就不相同呢？ s[0] 为什么就不能是最小重复子串。 如果 s[0] 和 s[1] 也相同，同时 s[0]s[1]与s[2]s[3]相同，s[2]s[3] 与 s[4]s[5]相同，s[4]s[5] 与 s[6]s[7] 相同，那么这个字符串就是有一个字符构成的字符串。 @@ -246,7 +246,7 @@ p2 = p1，p3 = p2 即： p1 = p2 = p3 或者说，自己举个例子，`aaaaaa`，这个字符串，他的最长相等前后缀是什么？ -同上以上推导，最长相等前后缀不包含的子串的长度只要被 字符串s的长度整除，就是一定是最小重复子串。 +同上以上推导，最长相等前后缀不包含的子串的长度只要被 字符串s的长度整除，最长相等前后缀不包含的子串一定是最小重复子串。 ---------------- @@ -267,7 +267,7 @@ p2 = p1，p3 = p2 即： p1 = p2 = p3 以上推导，可以得出 s[0],s[1],s[2] 与 s[3],s[4],s[5] 相同，s[3]s[4] 与 s[6]s[7]相同。 -那么 最长相等前后缀不包含的子串的长度 不被 字符串s的长度整除 ，就不是s的重复子串 +那么 最长相等前后缀不包含的子串的长度 不被 字符串s的长度整除 ，最长相等前后缀不包含的子串就不是s的重复子串 ----------- @@ -277,7 +277,7 @@ p2 = p1，p3 = p2 即： p1 = p2 = p3 在必要条件，这个是 显而易见的，都已经假设 最长相等前后缀不包含的子串 是 s的最小重复子串了，那s必然是重复子串。 -关键是需要证明， 字符串s的最长相等前后缀不包含的子串 什么时候才是 s最小重复子串。 +**关键是需要证明， 字符串s的最长相等前后缀不包含的子串 什么时候才是 s最小重复子串**。 同上我们证明了，当 最长相等前后缀不包含的子串的长度 可以被 字符串s的长度整除，那么不包含的子串 就是s的最小重复子串。 @@ -312,7 +312,7 @@ next 数组记录的就是最长相同前后缀（ [字符串：KMP算法精讲] 4可以被 12(字符串的长度) 整除，所以说明有重复的子字符串（asdf）。 -### 打码实现 +### 代码实现 C++代码如下：（这里使用了前缀表统一减一的实现方式） diff --git a/problems/0518.零钱兑换II.md b/problems/0518.零钱兑换II.md index 255912d6..360da582 100644 --- a/problems/0518.零钱兑换II.md +++ b/problems/0518.零钱兑换II.md @@ -168,23 +168,43 @@ for (int j = 0; j <= amount; j++) { // 遍历背包容量 class Solution { public: int change(int amount, vector& coins) { - vector dp(amount + 1, 0); - dp[0] = 1; + vector dp(amount + 1, 0); // 防止相加数据超int + dp[0] = 1; // 只有一种方式达到0 for (int i = 0; i < coins.size(); i++) { // 遍历物品 for (int j = coins[i]; j <= amount; j++) { // 遍历背包 dp[j] += dp[j - coins[i]]; } } - return dp[amount]; + return dp[amount]; // 返回组合数 } }; ``` +C++测试用例有两个数相加超过int的数据，所以需要在if里加上dp[i] < INT_MAX - dp[i - num]。 + * 时间复杂度: O(mn)，其中 m 是amount，n 是 coins 的长度 * 空间复杂度: O(m) +为了防止相加的数据 超int 也可以这么写： + +```CPP +class Solution { +public: + int change(int amount, vector& coins) { + vector dp(amount + 1, 0); + dp[0] = 1; // 只有一种方式达到0 + for (int i = 0; i < coins.size(); i++) { // 遍历物品 + for (int j = coins[i]; j <= amount; j++) { // 遍历背包 + if (dp[j] < INT_MAX - dp[j - coins[i]]) { //防止相加数据超int + dp[j] += dp[j - coins[i]]; + } + } + } + return dp[amount]; // 返回组合数 + } +}; +``` -是不是发现代码如此精简 ## 总结 From 8f692f8ff4545337e9df4790fa2cdb7f9a07c219 Mon Sep 17 00:00:00 2001 From: 01geek Date: Wed, 9 Oct 2024 17:36:56 +0800 Subject: [PATCH 54/64] =?UTF-8?q?=E7=BA=A0=E6=AD=A3=200150.=E9=80=86?= =?UTF-8?q?=E6=B3=A2=E5=85=B0=E8=A1=A8=E8=BE=BE=E5=BC=8F=E6=B1=82=E5=80=BC?= =?UTF-8?q?.md=20=E7=9A=84c++=E4=BB=A3=E7=A0=81=EF=BC=8C=E5=B0=86=E7=AC=AC?= =?UTF-8?q?21=E8=A1=8C=E4=BB=A3=E7=A0=81=E7=9A=84int=E4=BF=AE=E6=94=B9?= =?UTF-8?q?=E4=B8=BAlong=20long=EF=BC=8C=E5=8E=9F=E5=9B=A0=EF=BC=9A?= =?UTF-8?q?=E6=97=A2=E7=84=B6=E5=89=8D=E9=9D=A2=E8=80=83=E8=99=91=E5=88=B0?= =?UTF-8?q?=E5=8F=AF=E8=83=BD=E6=95=B0=E6=8D=AE=E5=80=BC=E5=A4=A7=E8=80=8C?= =?UTF-8?q?=E5=B0=86int=E6=94=B9=E4=B8=BA=E4=BA=86long=20long=EF=BC=8C?= =?UTF-8?q?=E9=82=A3=E4=B9=88=E6=9C=80=E5=90=8E=E7=9A=84RPN=E8=BF=90?= =?UTF-8?q?=E7=AE=97=E7=BB=93=E6=9E=9C=E4=B9=9F=E5=8F=AF=E8=83=BD=E8=B6=85?= =?UTF-8?q?=E8=BF=87int=E8=83=BD=E8=A1=A8=E7=A4=BA=E7=9A=84=E8=8C=83?= =?UTF-8?q?=E5=9B=B4?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0150.逆波兰表达式求值.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0150.逆波兰表达式求值.md b/problems/0150.逆波兰表达式求值.md index 48e99c5b..7d4031d7 100644 --- a/problems/0150.逆波兰表达式求值.md +++ b/problems/0150.逆波兰表达式求值.md @@ -108,7 +108,7 @@ public: } } - int result = st.top(); + long long result = st.top(); st.pop(); // 把栈里最后一个元素弹出（其实不弹出也没事） return result; } From a45758ef5f629807e248ea87407611bcbf42ed88 Mon Sep 17 00:00:00 2001 From: iYaovo <2293257241@qq.com> Date: Fri, 11 Oct 2024 10:13:18 +0800 Subject: [PATCH 55/64] =?UTF-8?q?docs:=E8=A1=A5=E5=85=850046Java=E4=BB=A3?= =?UTF-8?q?=E7=A0=81=E7=9A=84return=E8=AF=AD=E5=8F=A5?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0046.全排列.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/0046.全排列.md b/problems/0046.全排列.md index 1ef80a14..ab7af3ba 100644 --- a/problems/0046.全排列.md +++ b/problems/0046.全排列.md @@ -201,6 +201,7 @@ class Solution { public void backtrack(int[] nums, LinkedList path) { if (path.size() == nums.length) { result.add(new ArrayList<>(path)); + return; } for (int i =0; i < nums.length; i++) { // 如果path中已有，则跳过 @@ -524,3 +525,4 @@ public class Solution + From 19a6a4e369277cd192f6b149b008ce2d0cba23a4 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Lane=20Zhang=20=28=E5=BC=A0=E5=81=A5=29?= Date: Mon, 14 Oct 2024 10:00:24 +0800 Subject: [PATCH 56/64] =?UTF-8?q?Update=200202.=E5=BF=AB=E4=B9=90=E6=95=B0?= =?UTF-8?q?.md=20=E5=A2=9E=E5=8A=A0=20Ruby=20=E7=89=88=E6=9C=AC=E5=AE=9E?= =?UTF-8?q?=E7=8E=B0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0202.快乐数.md | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md index 409a7471..39cb39fa 100644 --- a/problems/0202.快乐数.md +++ b/problems/0202.快乐数.md @@ -534,6 +534,30 @@ public class Solution { } ``` +### Ruby： + +```ruby +# @param {Integer} n +# @return {Boolean} +def is_happy(n) + @occurred_nums = Set.new + + while true + n = next_value(n) + + return true if n == 1 + + return false if @occurred_nums.include?(n) + + @occurred_nums << n + end +end + +def next_value(n) + n.to_s.chars.sum { |char| char.to_i ** 2 } +end +``` +

From 0be8c49ba72d9e3e6ccdc4b76949a07ae128b7d5 Mon Sep 17 00:00:00 2001 From: markwang Date: Tue, 15 Oct 2024 09:49:57 +0800 Subject: [PATCH 57/64] =?UTF-8?q?0714.=E4=B9=B0=E5=8D=96=E8=82=A1=E7=A5=A8?= =?UTF-8?q?=E7=9A=84=E6=9C=80=E4=BD=B3=E6=97=B6=E6=9C=BA=E5=90=AB=E6=89=8B?= =?UTF-8?q?=E7=BB=AD=E8=B4=B9=E6=96=87=E5=AD=97=E5=8B=98=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0714.买卖股票的最佳时机含手续费.md | 4 ++-- problems/0714.买卖股票的最佳时机含手续费（动态规划）.md | 4 ++-- 2 files changed, 4 insertions(+), 4 deletions(-) diff --git a/problems/0714.买卖股票的最佳时机含手续费.md b/problems/0714.买卖股票的最佳时机含手续费.md index 88b03d9d..8a2d976b 100644 --- a/problems/0714.买卖股票的最佳时机含手续费.md +++ b/problems/0714.买卖股票的最佳时机含手续费.md @@ -43,7 +43,7 @@ 在[贪心算法：122.买卖股票的最佳时机II](https://programmercarl.com/0122.买卖股票的最佳时机II.html)中使用贪心策略不用关心具体什么时候买卖，只要收集每天的正利润，最后稳稳的就是最大利润了。 -而本题有了手续费，就要关系什么时候买卖了，因为计算所获得利润，需要考虑买卖利润可能不足以手续费的情况。 +而本题有了手续费，就要关心什么时候买卖了，因为计算所获得利润，需要考虑买卖利润可能不足以扣减手续费的情况。 如果使用贪心策略，就是最低值买，最高值（如果算上手续费还盈利）就卖。 @@ -122,7 +122,7 @@ public: * 时间复杂度：O(n) * 空间复杂度：O(n) -当然可以对空间经行优化，因为当前状态只是依赖前一个状态。 +当然可以对空间进行优化，因为当前状态只是依赖前一个状态。 C++ 代码如下： diff --git a/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md b/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md index b0e8b141..86e3e88f 100644 --- a/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md +++ b/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md @@ -46,7 +46,7 @@ * 时间复杂度：O(n) * 空间复杂度：O(1) -本题使用贪心算法并不好理解，也很容易出错，那么我们再来看看是使用动规的方法如何解题。 +本题使用贪心算法并不好理解，也很容易出错，那么我们再来看看使用动规的方法如何解题。 相对于[动态规划：122.买卖股票的最佳时机II](https://programmercarl.com/0122.买卖股票的最佳时机II（动态规划）.html)，本题只需要在计算卖出操作的时候减去手续费就可以了，代码几乎是一样的。 @@ -54,7 +54,7 @@ 这里重申一下dp数组的含义： -dp[i][0] 表示第i天持有股票所省最多现金。 +dp[i][0] 表示第i天持有股票所得最多现金。 dp[i][1] 表示第i天不持有股票所得最多现金 From abd3593c67a551bbe01136a42defabf219bb55be Mon Sep 17 00:00:00 2001 From: Lane Zhang Date: Tue, 15 Oct 2024 13:30:09 +0800 Subject: [PATCH 58/64] =?UTF-8?q?0454.=E5=9B=9B=E6=95=B0=E7=9B=B8=E5=8A=A0?= =?UTF-8?q?II.html=20=E5=A2=9E=E5=8A=A0Ruby=E7=89=88=E6=9C=AC=E5=AE=9E?= =?UTF-8?q?=E7=8E=B0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0454.四数相加II.md | 38 +++++++++++++++++++++++++++++++++++++ 1 file changed, 38 insertions(+) diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md index 6231c22b..af19f5f7 100644 --- a/problems/0454.四数相加II.md +++ b/problems/0454.四数相加II.md @@ -488,6 +488,44 @@ int fourSumCount(int* nums1, int nums1Size, int* nums2, int nums2Size, int* nums } ``` +### Ruby: + +```ruby +# @param {Integer[]} nums1 +# @param {Integer[]} nums2 +# @param {Integer[]} nums3 +# @param {Integer[]} nums4 +# @return {Integer} +# 新思路：和版主的思路基本相同，只是对后面两个数组的二重循环，用一个方法调用外加一重循环替代，简化了一点。 +# 简单的说，就是把四数和变成了两个两数和的统计（结果放到两个 hash 中），然后再来一次两数和为0. +# 把四个数分成两组两个数，然后分别计算每组可能的和情况，分别存入 hash 中，key 是 和，value 是 数量； +# 最后，得到的两个 hash 只需要遍历一次，符合和为零的 value 相乘并加总。 +def four_sum_count(nums1, nums2, nums3, nums4) + num_to_count_1 = two_sum_mapping(nums1, nums2) + num_to_count_2 = two_sum_mapping(nums3, nums4) + + count_sum = 0 + + num_to_count_1.each do |num, count| + count_sum += num_to_count_2[-num] * count # 反查另一个 hash，看有没有匹配的，没有的话，hash 默认值为 0，不影响加总；有匹配的，乘积就是可能的情况 + end + + count_sum +end + +def two_sum_mapping(nums1, nums2) + num_to_count = Hash.new(0) + + nums1.each do |num1| + nums2.each do |nums2| + num_to_count[num1 + nums2] += 1 # 统计和为 num1 + nums2 的有几个 + end + end + + num_to_count +end +``` +

From 83565ed41768333098fe78694f9990b280f64879 Mon Sep 17 00:00:00 2001 From: Leehouc <152672308+Leehouc@users.noreply.github.com> Date: Wed, 23 Oct 2024 20:28:57 +0800 Subject: [PATCH 59/64] =?UTF-8?q?=E9=94=99=E5=AD=970450.=E5=88=A0=E9=99=A4?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84?= =?UTF-8?q?=E8=8A=82=E7=82=B9.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0450.删除二叉搜索树中的节点.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0450.删除二叉搜索树中的节点.md b/problems/0450.删除二叉搜索树中的节点.md index f6057f44..cab9880a 100644 --- a/problems/0450.删除二叉搜索树中的节点.md +++ b/problems/0450.删除二叉搜索树中的节点.md @@ -26,7 +26,7 @@ ## 算法公开课 -**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[调整二叉树的结构最难！| LeetCode：450.删除二叉搜索树中的节点](https://www.bilibili.com/video/BV1tP41177us?share_source=copy_web)，相信结合视频在看本篇题解，更有助于大家对本题的理解**。 +**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[调整二叉树的结构最难！| LeetCode：450.删除二叉搜索树中的节点](https://www.bilibili.com/video/BV1tP41177us?share_source=copy_web)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。 ## 思路 From b9a060f195f704a09b7acf50e3ff36cb090471d0 Mon Sep 17 00:00:00 2001 From: Captainzw <80784797+Captain-zhangw@users.noreply.github.com> Date: Sat, 26 Oct 2024 20:16:14 +0800 Subject: [PATCH 60/64] =?UTF-8?q?=E4=BF=AE=E6=94=B9=20README.md=E4=B8=AD?= =?UTF-8?q?=20=E2=80=9Cbellman=5Fford=E4=B9=8B=E5=8D=95=E6=BA=90=E6=9C=89?= =?UTF-8?q?=E9=99=90=E6=9C=80=E7=9F=AD=E8=B7=AF=E2=80=9D=E7=9A=84=E9=94=99?= =?UTF-8?q?=E8=AF=AF=E9=93=BE=E6=8E=A5?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- README.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/README.md b/README.md index 3bb575d1..d1f4f3a8 100644 --- a/README.md +++ b/README.md @@ -400,7 +400,7 @@ 24. [图论：Bellman_ford 算法](./problems/kamacoder/0094.城市间货物运输I.md) 25. [图论：Bellman_ford 队列优化算法（又名SPFA）](./problems/kamacoder/0094.城市间货物运输I-SPFA.md) 26. [图论：Bellman_ford之判断负权回路](./problems/kamacoder/0095.城市间货物运输II.md) -27. [图论：Bellman_ford之单源有限最短路](./problems/kamacoder/0095.城市间货物运输II.md) +27. [图论：Bellman_ford之单源有限最短路](./problems/kamacoder/0096.城市间货物运输III.md) 28. [图论：Floyd 算法](./problems/kamacoder/0097.小明逛公园.md) 29. [图论：A * 算法](./problems/kamacoder/0126.骑士的攻击astar.md) 30. [图论：最短路算法总结篇](./problems/kamacoder/最短路问题总结篇.md) From 74909977cd13549da4368f09b550da2474f1c807 Mon Sep 17 00:00:00 2001 From: Captainzw <80784797+Captain-zhangw@users.noreply.github.com> Date: Sat, 26 Oct 2024 20:31:53 +0800 Subject: [PATCH 61/64] =?UTF-8?q?=E4=BF=AE=E6=94=B9=200096.=E5=9F=8E?= =?UTF-8?q?=E5=B8=82=E9=97=B4=E8=B4=A7=E7=89=A9=E8=BF=90=E8=BE=93III.md=20?= =?UTF-8?q?=E4=B8=AD=E7=9A=84=E9=94=99=E8=AF=AF=E9=93=BE=E6=8E=A5?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0096.城市间货物运输III.md | 16 ++++++++-------- 1 file changed, 8 insertions(+), 8 deletions(-) diff --git a/problems/kamacoder/0096.城市间货物运输III.md b/problems/kamacoder/0096.城市间货物运输III.md index 567a1d87..1facf7b2 100644 --- a/problems/kamacoder/0096.城市间货物运输III.md +++ b/problems/kamacoder/0096.城市间货物运输III.md @@ -51,15 +51,15 @@ ## 思路 -本题为单源有限最短路问题，同样是 [kama94.城市间货物运输I](./kama94.城市间货物运输I.md) 延伸题目。 +本题为单源有限最短路问题，同样是 [kama94.城市间货物运输I](./0094.城市间货物运输I.md) 延伸题目。 注意题目中描述是 **最多经过 k 个城市的条件下，而不是一定经过k个城市，也可以经过的城市数量比k小，但要最短的路径**。 -在 [kama94.城市间货物运输I](./kama94.城市间货物运输I.md) 中我们讲了：**对所有边松弛一次，相当于计算 起点到达 与起点一条边相连的节点 的最短距离**。 +在 [kama94.城市间货物运输I](./0094.城市间货物运输I.md) 中我们讲了：**对所有边松弛一次，相当于计算 起点到达 与起点一条边相连的节点 的最短距离**。 节点数量为n，起点到终点，最多是 n-1 条边相连。 那么对所有边松弛 n-1 次 就一定能得到 起点到达 终点的最短距离。 -（如果对以上讲解看不懂，建议详看 [kama94.城市间货物运输I](./kama94.城市间货物运输I.md) ） +（如果对以上讲解看不懂，建议详看 [kama94.城市间货物运输I](./0094.城市间货物运输I.md) ） 本题是最多经过 k 个城市， 那么是 k + 1条边相连的节点。 这里可能有录友想不懂为什么是k + 1，来看这个图： @@ -71,7 +71,7 @@ 对所有边松弛一次，相当于计算 起点到达 与起点一条边相连的节点 的最短距离，那么对所有边松弛 k + 1次，就是求 起点到达 与起点k + 1条边相连的节点的 最短距离。 -**注意**： 本题是 [kama94.城市间货物运输I](./kama94.城市间货物运输I.md) 的拓展题，如果对 bellman_ford 没有深入了解，强烈建议先看 [kama94.城市间货物运输I](./kama94.城市间货物运输I.md) 再做本题。 +**注意**： 本题是 [kama94.城市间货物运输I](./0094.城市间货物运输I.md) 的拓展题，如果对 bellman_ford 没有深入了解，强烈建议先看 [kama94.城市间货物运输I](./0094.城市间货物运输I.md) 再做本题。 理解以上内容，其实本题代码就很容易了，bellman_ford 标准写法是松弛 n-1 次，本题就松弛 k + 1次就好。 @@ -366,19 +366,19 @@ int main() { ## 拓展二（本题本质） -那么前面讲解过的 [94.城市间货物运输I](./kama94.城市间货物运输I.md) 和 [95.城市间货物运输II](./kama95.城市间货物运输II.md) 也是bellman_ford经典算法，也没使用 minDist_copy，怎么就没问题呢？ +那么前面讲解过的 [94.城市间货物运输I](./0094.城市间货物运输I.md) 和 [95.城市间货物运输II](./0095.城市间货物运输II.md) 也是bellman_ford经典算法，也没使用 minDist_copy，怎么就没问题呢？ > 如果没看过我上面这两篇讲解的话，建议详细学习上面两篇，再看我下面讲的区别，否则容易看不懂。 -[94.城市间货物运输I](./kama94.城市间货物运输I.md)， 是没有 负权回路的，那么 多松弛多少次，对结果都没有影响。 +[94.城市间货物运输I](./0094.城市间货物运输I.md)， 是没有 负权回路的，那么 多松弛多少次，对结果都没有影响。 求 节点1 到 节点n 的最短路径，松弛n-1 次就够了，松弛 大于 n-1次，结果也不会变。 那么在对所有边进行第一次松弛的时候，如果基于 本次计算的 minDist 来计算 minDist （相当于多做松弛了），也是对最终结果没影响。 -[95.城市间货物运输II](./kama95.城市间货物运输II.md) 是判断是否有 负权回路，一旦有负权回路， 对所有边松弛 n-1 次以后，在做松弛 minDist 数值一定会变，根据这一点来判断是否有负权回路。 +[95.城市间货物运输II](./0095.城市间货物运输II.md) 是判断是否有 负权回路，一旦有负权回路， 对所有边松弛 n-1 次以后，在做松弛 minDist 数值一定会变，根据这一点来判断是否有负权回路。 -所以，[95.城市间货物运输II](./kama95.城市间货物运输II.md) 只需要判断minDist数值变化了就行，而 minDist 的数值对不对，并不是我们关心的。 +所以，[95.城市间货物运输II](./0095.城市间货物运输II.md) 只需要判断minDist数值变化了就行，而 minDist 的数值对不对，并不是我们关心的。 那么本题 为什么计算minDist 一定要基于上次 的 minDist 数值。 From 7cb657dc508465362d74b6133bd060c42eddf610 Mon Sep 17 00:00:00 2001 From: Captainzw <80784797+Captain-zhangw@users.noreply.github.com> Date: Sat, 26 Oct 2024 20:33:16 +0800 Subject: [PATCH 62/64] =?UTF-8?q?=E4=BF=AE=E6=94=B9=200095.=E5=9F=8E?= =?UTF-8?q?=E5=B8=82=E9=97=B4=E8=B4=A7=E7=89=A9=E8=BF=90=E8=BE=93II.md=20?= =?UTF-8?q?=E4=B8=AD=E7=9A=84=E9=94=99=E8=AF=AF=E9=93=BE=E6=8E=A5?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0095.城市间货物运输II.md | 12 ++++++------ 1 file changed, 6 insertions(+), 6 deletions(-) diff --git a/problems/kamacoder/0095.城市间货物运输II.md b/problems/kamacoder/0095.城市间货物运输II.md index edfd52e0..ac6ccf3c 100644 --- a/problems/kamacoder/0095.城市间货物运输II.md +++ b/problems/kamacoder/0095.城市间货物运输II.md @@ -54,7 +54,7 @@ circle ## 思路 -本题是 [kama94.城市间货物运输I](./kama94.城市间货物运输I.md) 延伸题目。 +本题是 [kama94.城市间货物运输I](./0094.城市间货物运输I.md) 延伸题目。 本题是要我们判断 负权回路，也就是图中出现环且环上的边总权值为负数。 @@ -64,7 +64,7 @@ circle 接下来我们来看 如何使用 bellman_ford 算法来判断 负权回路。 -在 [kama94.城市间货物运输I](./kama94.城市间货物运输I.md) 中 我们讲了 bellman_ford 算法的核心就是一句话：对 所有边 进行 n-1 次松弛。 同时文中的 【拓展】部分， 我们也讲了 松弛n次以上 会怎么样？ +在 [kama94.城市间货物运输I](./0094.城市间货物运输I.md) 中 我们讲了 bellman_ford 算法的核心就是一句话：对 所有边 进行 n-1 次松弛。 同时文中的 【拓展】部分， 我们也讲了 松弛n次以上 会怎么样？ 在没有负权回路的图中，松弛 n 次以上 ，结果不会有变化。 @@ -72,7 +72,7 @@ circle 那么每松弛一次，都会更新最短路径，所以结果会一直有变化。 -（如果对于 bellman_ford 不了解的录友，建议详细看这里：[kama94.城市间货物运输I](./kama94.城市间货物运输I.md)） +（如果对于 bellman_ford 不了解的录友，建议详细看这里：[kama94.城市间货物运输I](./0094.城市间货物运输I.md)） 以上为理论分析，接下来我们再画图举例。 @@ -94,13 +94,13 @@ circle 如果在负权回路多绕两圈，三圈，无穷圈，那么我们的总成本就会无限小， 如果要求最小成本的话，你会发现本题就无解了。 -在 bellman_ford 算法中，松弛 n-1 次所有的边 就可以求得 起点到任何节点的最短路径，松弛 n 次以上，minDist数组（记录起到到其他节点的最短距离）中的结果也不会有改变 （如果对 bellman_ford 算法 不了解，也不知道 minDist 是什么，建议详看上篇讲解[kama94.城市间货物运输I](./kama94.城市间货物运输I.md)） +在 bellman_ford 算法中，松弛 n-1 次所有的边 就可以求得 起点到任何节点的最短路径，松弛 n 次以上，minDist数组（记录起到到其他节点的最短距离）中的结果也不会有改变 （如果对 bellman_ford 算法 不了解，也不知道 minDist 是什么，建议详看上篇讲解[kama94.城市间货物运输I](./0094.城市间货物运输I.md)） 而本题有负权回路的情况下，一直都会有更短的最短路，所以 松弛 第n次，minDist数组 也会发生改变。 -那么解决本题的 核心思路，就是在 [kama94.城市间货物运输I](./kama94.城市间货物运输I.md) 的基础上，再多松弛一次，看minDist数组 是否发生变化。 +那么解决本题的 核心思路，就是在 [kama94.城市间货物运输I](./0094.城市间货物运输I.md) 的基础上，再多松弛一次，看minDist数组 是否发生变化。 -代码和 [kama94.城市间货物运输I](./kama94.城市间货物运输I.md) 基本是一样的，如下：（关键地方已注释） +代码和 [kama94.城市间货物运输I](./0094.城市间货物运输I.md) 基本是一样的，如下：（关键地方已注释） ```CPP #include From 85241be1ea8a07b6449b2ef24d4abfacf4211bc1 Mon Sep 17 00:00:00 2001 From: DengSchoo <46556279+DengSchoo@users.noreply.github.com> Date: Tue, 5 Nov 2024 17:41:16 +0800 Subject: [PATCH 63/64] chore: create .gitignore file create a .gitignore file to avoid uploading useless files, such as ".DS_Store". --- .gitignore | 7 +++++++ 1 file changed, 7 insertions(+) create mode 100644 .gitignore diff --git a/.gitignore b/.gitignore new file mode 100644 index 00000000..4bd29f75 --- /dev/null +++ b/.gitignore @@ -0,0 +1,7 @@ +.idea/ +.DS_Store +.vscode +.temp +.cache +*.iml +__pycache__ From 55e8b33e1d9f9236b939d2be7a0160eecfb2beb8 Mon Sep 17 00:00:00 2001 From: DengSchoo <46556279+DengSchoo@users.noreply.github.com> Date: Tue, 5 Nov 2024 17:42:16 +0800 Subject: [PATCH 64/64] chore: remove useless file ".DS_Store" --- .DS_Store | Bin 6148 -> 0 bytes 1 file changed, 0 insertions(+), 0 deletions(-) delete mode 100644 .DS_Store diff --git a/.DS_Store b/.DS_Store deleted file mode 100644 index f7ce0fd5038358c7a0df77b968ed993daff0639c..0000000000000000000000000000000000000000 GIT binary patch literal 0 HcmV?d00001 literal 6148 zcmeHKF-`+P3>-s>lW0;>?icvMDhe;i2T%y2IdUiw)YtKC#$F>OS}4*`z*zFmt?k*> zO>sU0u+`h{4p;&h=#KdCr8m#JcXpK#M~gE?JmCe$-SKhks=p67cfczSn6SkOe>+Vh z5ABoho5Oyb_Rq|DUM2;kfE17dQa}n^q(HXM;pQTiilu-Q_zw#B_o2`oYvSB6J{??Q z1RyRL4&!so62#&GVojVIGD5RPB{r(%h+&P+c*(k&I5%u`SlrBW@@C5k#o~6xOO(TE zL#5>MJ5G)LxE2gD;uo<