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feat: add solutions to lcci problems: No.08.06~08.09 (#4068)

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2
lcci/08.06.Hanota/README.md
View File

@ -91,7 +91,7 @@ class Solution {
class Solution {
public:
void hanota(vector<int>& A, vector<int>& B, vector<int>& C) {
function<void(int, vector<int>&, vector<int>&, vector<int>&)> dfs = [&](int n, vector<int>& a, vector<int>& b, vector<int>& c) {
auto dfs = [&](this auto&& dfs, int n, vector<int>& a, vector<int>& b, vector<int>& c) {
if (n == 1) {
c.push_back(a.back());
a.pop_back();

2
lcci/08.06.Hanota/README_EN.md
View File

@ -98,7 +98,7 @@ class Solution {
class Solution {
public:
void hanota(vector<int>& A, vector<int>& B, vector<int>& C) {
function<void(int, vector<int>&, vector<int>&, vector<int>&)> dfs = [&](int n, vector<int>& a, vector<int>& b, vector<int>& c) {
auto dfs = [&](this auto&& dfs, int n, vector<int>& a, vector<int>& b, vector<int>& c) {
if (n == 1) {
c.push_back(a.back());
a.pop_back();

4
lcci/08.06.Hanota/Solution.cpp
View File

@ -1,7 +1,7 @@
class Solution {
public:
void hanota(vector<int>& A, vector<int>& B, vector<int>& C) {
function<void(int, vector<int>&, vector<int>&, vector<int>&)> dfs = [&](int n, vector<int>& a, vector<int>& b, vector<int>& c) {
auto dfs = [&](this auto&& dfs, int n, vector<int>& a, vector<int>& b, vector<int>& c) {
if (n == 1) {
c.push_back(a.back());
a.pop_back();
@ -14,4 +14,4 @@ public:
};
dfs(A.size(), A, B, C);
}
};
};

138
lcci/08.07.Permutation I/README.md
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@ -45,7 +45,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcci/08.07.Permutation%20I
### 方法一：DFS（回溯）
我们设计一个函数 $dfs(i)$ 表示已经填完了前 $i$ 个位置，现在需要填第 $i+1$ 个位置。枚举所有可能的字符，如果这个字符没有被填过，就填入这个字符，然后继续填下一个位置，直到填完所有的位置。
我们设计一个函数 $\textit{dfs}(i)$ 表示已经填完了前 $i$ 个位置，现在需要填第 $i+1$ 个位置。枚举所有可能的字符，如果这个字符没有被填过，就填入这个字符，然后继续填下一个位置，直到填完所有的位置。
时间复杂度 $O(n \times n!)$，其中 $n$ 是字符串的长度。一共有 $n!$ 个排列，每个排列需要 $O(n)$ 的时间来构造。
@ -57,22 +57,20 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcci/08.07.Permutation%20I
class Solution:
def permutation(self, S: str) -> List[str]:
def dfs(i: int):
if i == n:
if i >= n:
ans.append("".join(t))
return
for j, c in enumerate(S):
if vis[j]:
continue
vis[j] = True
t.append(c)
dfs(i + 1)
t.pop()
vis[j] = False
if not vis[j]:
vis[j] = True
t[i] = c
dfs(i + 1)
vis[j] = False
ans = []
n = len(S)
vis = [False] * n
ans = []
t = []
t = list(S)
dfs(0)
return ans
```
@ -82,30 +80,31 @@ class Solution:
```java
class Solution {
private char[] s;
private boolean[] vis = new boolean['z' + 1];
private char[] t;
private boolean[] vis;
private List<String> ans = new ArrayList<>();
private StringBuilder t = new StringBuilder();
public String[] permutation(String S) {
s = S.toCharArray();
int n = s.length;
vis = new boolean[n];
t = new char[n];
dfs(0);
return ans.toArray(new String[0]);
}
private void dfs(int i) {
if (i == s.length) {
ans.add(t.toString());
if (i >= s.length) {
ans.add(new String(t));
return;
}
for (char c : s) {
if (vis[c]) {
continue;
for (int j = 0; j < s.length; ++j) {
if (!vis[j]) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[c] = true;
t.append(c);
dfs(i + 1);
t.deleteCharAt(t.length() - 1);
vis[c] = false;
}
}
}
@ -119,51 +118,49 @@ public:
vector<string> permutation(string S) {
int n = S.size();
vector<bool> vis(n);
string t = S;
vector<string> ans;
string t;
function<void(int)> dfs = [&](int i) {
auto dfs = [&](this auto&& dfs, int i) {
if (i >= n) {
ans.push_back(t);
ans.emplace_back(t);
return;
}
for (int j = 0; j < n; ++j) {
if (vis[j]) {
continue;
if (!vis[j]) {
vis[j] = true;
t[i] = S[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push_back(S[j]);
dfs(i + 1);
t.pop_back();
vis[j] = false;
}
};
dfs(0);
return ans;
}
};
```
#### Go
```go
func permutation(S string) (ans []string) {
t := []byte{}
vis := make([]bool, len(S))
t := []byte(S)
n := len(t)
vis := make([]bool, n)
var dfs func(int)
dfs = func(i int) {
if i >= len(S) {
if i >= n {
ans = append(ans, string(t))
return
}
for j := range S {
if vis[j] {
continue
if !vis[j] {
vis[j] = true
t[i] = S[j]
dfs(i + 1)
vis[j] = false
}
vis[j] = true
t = append(t, S[j])
dfs(i + 1)
t = t[:len(t)-1]
vis[j] = false
}
}
dfs(0)
@ -178,7 +175,7 @@ function permutation(S: string): string[] {
const n = S.length;
const vis: boolean[] = Array(n).fill(false);
const ans: string[] = [];
const t: string[] = [];
const t: string[] = Array(n).fill('');
const dfs = (i: number) => {
if (i >= n) {
ans.push(t.join(''));
@ -189,9 +186,8 @@ function permutation(S: string): string[] {
continue;
}
vis[j] = true;
t.push(S[j]);
t[i] = S[j];
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
@ -211,7 +207,7 @@ var permutation = function (S) {
const n = S.length;
const vis = Array(n).fill(false);
const ans = [];
const t = [];
const t = Array(n).fill('');
const dfs = i => {
if (i >= n) {
ans.push(t.join(''));
@ -222,9 +218,8 @@ var permutation = function (S) {
continue;
}
vis[j] = true;
t.push(S[j]);
t[i] = S[j];
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
@ -237,34 +232,31 @@ var permutation = function (S) {
```swift
class Solution {
private var s: [Character] = []
private var vis: [Bool] = Array(repeating: false, count: 128)
private var ans: [String] = []
private var t: String = ""
func permutation(_ S: String) -> [String] {
s = Array(S)
var ans: [String] = []
let s = Array(S)
var t = s
var vis = Array(repeating: false, count: s.count)
let n = s.count
func dfs(_ i: Int) {
if i >= n {
ans.append(String(t))
return
}
for j in 0..<n {
if !vis[j] {
vis[j] = true
t[i] = s[j]
dfs(i + 1)
vis[j] = false
}
}
}
dfs(0)
return ans
}
private func dfs(_ i: Int) {
if i == s.count {
ans.append(t)
return
}
for c in s {
let index = Int(c.asciiValue!)
if vis[index] {
continue
}
vis[index] = true
t.append(c)
dfs(i + 1)
t.removeLast()
vis[index] = false
}
}
}
```

137
lcci/08.07.Permutation I/README_EN.md
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@ -51,7 +51,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcci/08.07.Permutation%20I
### Solution 1: DFS (Backtracking)
We design a function $dfs(i)$ to represent that the first $i$ positions have been filled, and now the $i+1$ position needs to be filled. We enumerate all possible characters, if this character has not been filled, we fill in this character, and then continue to fill the next position until all positions are filled.
We design a function $\textit{dfs}(i)$ to represent that the first $i$ positions have been filled, and now we need to fill the $(i+1)$-th position. Enumerate all possible characters, and if the character has not been used, fill in this character and continue to fill the next position until all positions are filled.
The time complexity is $O(n \times n!)$, where $n$ is the length of the string. There are $n!$ permutations in total, and each permutation takes $O(n)$ time to construct.
@ -63,22 +63,20 @@ The time complexity is $O(n \times n!)$, where $n$ is the length of the string.
class Solution:
def permutation(self, S: str) -> List[str]:
def dfs(i: int):
if i == n:
if i >= n:
ans.append("".join(t))
return
for j, c in enumerate(S):
if vis[j]:
continue
vis[j] = True
t.append(c)
dfs(i + 1)
t.pop()
vis[j] = False
if not vis[j]:
vis[j] = True
t[i] = c
dfs(i + 1)
vis[j] = False
ans = []
n = len(S)
vis = [False] * n
ans = []
t = []
t = list(S)
dfs(0)
return ans
```
@ -88,30 +86,31 @@ class Solution:
```java
class Solution {
private char[] s;
private boolean[] vis = new boolean['z' + 1];
private char[] t;
private boolean[] vis;
private List<String> ans = new ArrayList<>();
private StringBuilder t = new StringBuilder();
public String[] permutation(String S) {
s = S.toCharArray();
int n = s.length;
vis = new boolean[n];
t = new char[n];
dfs(0);
return ans.toArray(new String[0]);
}
private void dfs(int i) {
if (i == s.length) {
ans.add(t.toString());
if (i >= s.length) {
ans.add(new String(t));
return;
}
for (char c : s) {
if (vis[c]) {
continue;
for (int j = 0; j < s.length; ++j) {
if (!vis[j]) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[c] = true;
t.append(c);
dfs(i + 1);
t.deleteCharAt(t.length() - 1);
vis[c] = false;
}
}
}
@ -125,22 +124,20 @@ public:
vector<string> permutation(string S) {
int n = S.size();
vector<bool> vis(n);
string t = S;
vector<string> ans;
string t;
function<void(int)> dfs = [&](int i) {
auto dfs = [&](this auto&& dfs, int i) {
if (i >= n) {
ans.push_back(t);
ans.emplace_back(t);
return;
}
for (int j = 0; j < n; ++j) {
if (vis[j]) {
continue;
if (!vis[j]) {
vis[j] = true;
t[i] = S[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push_back(S[j]);
dfs(i + 1);
t.pop_back();
vis[j] = false;
}
};
dfs(0);
@ -153,23 +150,22 @@ public:
```go
func permutation(S string) (ans []string) {
t := []byte{}
vis := make([]bool, len(S))
t := []byte(S)
n := len(t)
vis := make([]bool, n)
var dfs func(int)
dfs = func(i int) {
if i >= len(S) {
if i >= n {
ans = append(ans, string(t))
return
}
for j := range S {
if vis[j] {
continue
if !vis[j] {
vis[j] = true
t[i] = S[j]
dfs(i + 1)
vis[j] = false
}
vis[j] = true
t = append(t, S[j])
dfs(i + 1)
t = t[:len(t)-1]
vis[j] = false
}
}
dfs(0)
@ -184,7 +180,7 @@ function permutation(S: string): string[] {
const n = S.length;
const vis: boolean[] = Array(n).fill(false);
const ans: string[] = [];
const t: string[] = [];
const t: string[] = Array(n).fill('');
const dfs = (i: number) => {
if (i >= n) {
ans.push(t.join(''));
@ -195,9 +191,8 @@ function permutation(S: string): string[] {
continue;
}
vis[j] = true;
t.push(S[j]);
t[i] = S[j];
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
@ -217,7 +212,7 @@ var permutation = function (S) {
const n = S.length;
const vis = Array(n).fill(false);
const ans = [];
const t = [];
const t = Array(n).fill('');
const dfs = i => {
if (i >= n) {
ans.push(t.join(''));
@ -228,9 +223,8 @@ var permutation = function (S) {
continue;
}
vis[j] = true;
t.push(S[j]);
t[i] = S[j];
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
@ -243,34 +237,31 @@ var permutation = function (S) {
```swift
class Solution {
private var s: [Character] = []
private var vis: [Bool] = Array(repeating: false, count: 128)
private var ans: [String] = []
private var t: String = ""
func permutation(_ S: String) -> [String] {
s = Array(S)
var ans: [String] = []
let s = Array(S)
var t = s
var vis = Array(repeating: false, count: s.count)
let n = s.count
func dfs(_ i: Int) {
if i >= n {
ans.append(String(t))
return
}
for j in 0..<n {
if !vis[j] {
vis[j] = true
t[i] = s[j]
dfs(i + 1)
vis[j] = false
}
}
}
dfs(0)
return ans
}
private func dfs(_ i: Int) {
if i == s.count {
ans.append(t)
return
}
for c in s {
let index = Int(c.asciiValue!)
if vis[index] {
continue
}
vis[index] = true
t.append(c)
dfs(i + 1)
t.removeLast()
vis[index] = false
}
}
}
```

20
lcci/08.07.Permutation I/Solution.cpp
View File

@ -3,25 +3,23 @@ public:
vector<string> permutation(string S) {
int n = S.size();
vector<bool> vis(n);
string t = S;
vector<string> ans;
string t;
function<void(int)> dfs = [&](int i) {
auto dfs = [&](this auto&& dfs, int i) {
if (i >= n) {
ans.push_back(t);
ans.emplace_back(t);
return;
}
for (int j = 0; j < n; ++j) {
if (vis[j]) {
continue;
if (!vis[j]) {
vis[j] = true;
t[i] = S[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push_back(S[j]);
dfs(i + 1);
t.pop_back();
vis[j] = false;
}
};
dfs(0);
return ans;
}
};
};

21
lcci/08.07.Permutation I/Solution.go
View File

@ -1,23 +1,22 @@
func permutation(S string) (ans []string) {
t := []byte{}
vis := make([]bool, len(S))
t := []byte(S)
n := len(t)
vis := make([]bool, n)
var dfs func(int)
dfs = func(i int) {
if i >= len(S) {
if i >= n {
ans = append(ans, string(t))
return
}
for j := range S {
if vis[j] {
continue
if !vis[j] {
vis[j] = true
t[i] = S[j]
dfs(i + 1)
vis[j] = false
}
vis[j] = true
t = append(t, S[j])
dfs(i + 1)
t = t[:len(t)-1]
vis[j] = false
}
}
dfs(0)
return
}
}

27
lcci/08.07.Permutation I/Solution.java
View File

@ -1,29 +1,30 @@
class Solution {
private char[] s;
private boolean[] vis = new boolean['z' + 1];
private char[] t;
private boolean[] vis;
private List<String> ans = new ArrayList<>();
private StringBuilder t = new StringBuilder();
public String[] permutation(String S) {
s = S.toCharArray();
int n = s.length;
vis = new boolean[n];
t = new char[n];
dfs(0);
return ans.toArray(new String[0]);
}
private void dfs(int i) {
if (i == s.length) {
ans.add(t.toString());
if (i >= s.length) {
ans.add(new String(t));
return;
}
for (char c : s) {
if (vis[c]) {
continue;
for (int j = 0; j < s.length; ++j) {
if (!vis[j]) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[c] = true;
t.append(c);
dfs(i + 1);
t.deleteCharAt(t.length() - 1);
vis[c] = false;
}
}
}
}

5
lcci/08.07.Permutation I/Solution.js
View File

@ -6,7 +6,7 @@ var permutation = function (S) {
const n = S.length;
const vis = Array(n).fill(false);
const ans = [];
const t = [];
const t = Array(n).fill('');
const dfs = i => {
if (i >= n) {
ans.push(t.join(''));
@ -17,9 +17,8 @@ var permutation = function (S) {
continue;
}
vis[j] = true;
t.push(S[j]);
t[i] = S[j];
dfs(i + 1);
t.pop();
vis[j] = false;
}
};

18
lcci/08.07.Permutation I/Solution.py
View File

@ -1,21 +1,19 @@
class Solution:
def permutation(self, S: str) -> List[str]:
def dfs(i: int):
if i == n:
if i >= n:
ans.append("".join(t))
return
for j, c in enumerate(S):
if vis[j]:
continue
vis[j] = True
t.append(c)
dfs(i + 1)
t.pop()
vis[j] = False
if not vis[j]:
vis[j] = True
t[i] = c
dfs(i + 1)
vis[j] = False
ans = []
n = len(S)
vis = [False] * n
ans = []
t = []
t = list(S)
dfs(0)
return ans

45
lcci/08.07.Permutation I/Solution.swift
View File

@ -1,30 +1,27 @@
class Solution {
private var s: [Character] = []
private var vis: [Bool] = Array(repeating: false, count: 128)
private var ans: [String] = []
private var t: String = ""
func permutation(_ S: String) -> [String] {
s = Array(S)
var ans: [String] = []
let s = Array(S)
var t = s
var vis = Array(repeating: false, count: s.count)
let n = s.count
func dfs(_ i: Int) {
if i >= n {
ans.append(String(t))
return
}
for j in 0..<n {
if !vis[j] {
vis[j] = true
t[i] = s[j]
dfs(i + 1)
vis[j] = false
}
}
}
dfs(0)
return ans
}
private func dfs(_ i: Int) {
if i == s.count {
ans.append(t)
return
}
for c in s {
let index = Int(c.asciiValue!)
if vis[index] {
continue
}
vis[index] = true
t.append(c)
dfs(i + 1)
t.removeLast()
vis[index] = false
}
}
}

5
lcci/08.07.Permutation I/Solution.ts
View File

@ -2,7 +2,7 @@ function permutation(S: string): string[] {
const n = S.length;
const vis: boolean[] = Array(n).fill(false);
const ans: string[] = [];
const t: string[] = [];
const t: string[] = Array(n).fill('');
const dfs = (i: number) => {
if (i >= n) {
ans.push(t.join(''));
@ -13,9 +13,8 @@ function permutation(S: string): string[] {
continue;
}
vis[j] = true;
t.push(S[j]);
t[i] = S[j];
dfs(i + 1);
t.pop();
vis[j] = false;
}
};

203
lcci/08.08.Permutation II/README.md
View File

@ -39,12 +39,12 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcci/08.08.Permutation%20I
我们可以先对字符串按照字符进行排序，这样就可以将重复的字符放在一起，方便我们进行去重。
然后，我们设计一个函数 $dfs(i)$，表示当前需要填写第 $i$ 个位置的字符。函数的具体实现如下：
然后，我们设计一个函数 $\textit{dfs}(i)$，表示当前需要填写第 $i$ 个位置的字符。函数的具体实现如下：
- 如果 $i = n$，说明我们已经填写完毕，将当前排列加入答案数组中，然后返回。
- 否则，我们枚举第 $i$ 个位置的字符 $s[j]$，其中 $j$ 的范围是 $[0, n - 1]$。我们需要保证 $s[j]$ 没有被使用过，并且与前面枚举的字符不同，这样才能保证当前排列不重复。如果满足条件，我们就可以填写 $s[j]$，并继续递归地填写下一个位置，即调用 $dfs(i + 1)$。在递归调用结束后，我们需要将 $s[j]$ 标记为未使用，以便于进行后面的枚举。
- 否则，我们枚举第 $i$ 个位置的字符 $\textit{s}[j]$，其中 $j$ 的范围是 $[0, n - 1]$。我们需要保证 $\textit{s}[j]$ 没有被使用过，并且与前面枚举的字符不同，这样才能保证当前排列不重复。如果满足条件，我们就可以填写 $\textit{s}[j]$，并继续递归地填写下一个位置，即调用 $\textit{dfs}(i + 1)$。在递归调用结束后，我们需要将 $\textit{s}[j]$ 标记为未使用，以便于进行后面的枚举。
在主函数中，我们首先对字符串进行排序，然后调用 $dfs(0)$，即从第 $0$ 个位置开始填写，最终返回答案数组即可。
在主函数中，我们首先对字符串进行排序，然后调用 $\textit{dfs}(0)$，即从第 $0$ 个位置开始填写，最终返回答案数组即可。
时间复杂度 $O(n \times n!)$，空间复杂度 $O(n)$。其中 $n$ 是字符串 $s$ 的长度。需要进行 $n!$ 次枚举，每次枚举需要 $O(n)$ 的时间来判断是否重复。另外，我们需要一个标记数组来标记每个位置是否被使用过，因此空间复杂度为 $O(n)$。
@ -56,21 +56,20 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcci/08.08.Permutation%20I
class Solution:
def permutation(self, S: str) -> List[str]:
def dfs(i: int):
if i == n:
if i >= n:
ans.append("".join(t))
return
for j in range(n):
if vis[j] or (j and cs[j] == cs[j - 1] and not vis[j - 1]):
continue
t[i] = cs[j]
vis[j] = True
dfs(i + 1)
vis[j] = False
for j, c in enumerate(s):
if not vis[j] and (j == 0 or s[j] != s[j - 1] or vis[j - 1]):
vis[j] = True
t[i] = c
dfs(i + 1)
vis[j] = False
cs = sorted(S)
n = len(cs)
s = sorted(S)
ans = []
t = [None] * n
t = s[:]
n = len(s)
vis = [False] * n
dfs(0)
return ans
@ -80,35 +79,33 @@ class Solution:
```java
class Solution {
private int n;
private char[] cs;
private List<String> ans = new ArrayList<>();
private char[] s;
private char[] t;
private boolean[] vis;
private StringBuilder t = new StringBuilder();
private List<String> ans = new ArrayList<>();
public String[] permutation(String S) {
cs = S.toCharArray();
n = cs.length;
Arrays.sort(cs);
int n = S.length();
s = S.toCharArray();
Arrays.sort(s);
t = new char[n];
vis = new boolean[n];
dfs(0);
return ans.toArray(new String[0]);
}
private void dfs(int i) {
if (i == n) {
ans.add(t.toString());
if (i >= s.length) {
ans.add(new String(t));
return;
}
for (int j = 0; j < n; ++j) {
if (vis[j] || (j > 0 && !vis[j - 1] && cs[j] == cs[j - 1])) {
continue;
for (int j = 0; j < s.length; ++j) {
if (!vis[j] && (j == 0 || s[j] != s[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.append(cs[j]);
dfs(i + 1);
t.deleteCharAt(t.length() - 1);
vis[j] = false;
}
}
}
@ -120,26 +117,23 @@ class Solution {
class Solution {
public:
vector<string> permutation(string S) {
vector<char> cs(S.begin(), S.end());
sort(cs.begin(), cs.end());
int n = cs.size();
vector<string> ans;
ranges::sort(S);
string t = S;
int n = t.size();
vector<bool> vis(n);
string t;
function<void(int)> dfs = [&](int i) {
if (i == n) {
ans.push_back(t);
vector<string> ans;
auto dfs = [&](this auto&& dfs, int i) {
if (i >= n) {
ans.emplace_back(t);
return;
}
for (int j = 0; j < n; ++j) {
if (vis[j] || (j && !vis[j - 1] && cs[j] == cs[j - 1])) {
continue;
if (!vis[j] && (j == 0 || S[j] != S[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = S[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push_back(cs[j]);
dfs(i + 1);
t.pop_back();
vis[j] = false;
}
};
dfs(0);
@ -152,26 +146,23 @@ public:
```go
func permutation(S string) (ans []string) {
cs := []byte(S)
sort.Slice(cs, func(i, j int) bool { return cs[i] < cs[j] })
t := []byte{}
n := len(cs)
vis := make([]bool, n)
s := []byte(S)
sort.Slice(s, func(i, j int) bool { return s[i] < s[j] })
t := slices.Clone(s)
vis := make([]bool, len(s))
var dfs func(int)
dfs = func(i int) {
if i == n {
if i >= len(s) {
ans = append(ans, string(t))
return
}
for j := 0; j < n; j++ {
if vis[j] || (j > 0 && !vis[j-1] && cs[j] == cs[j-1]) {
continue
for j := range s {
if !vis[j] && (j == 0 || s[j] != s[j-1] || vis[j-1]) {
vis[j] = true
t[i] = s[j]
dfs(i + 1)
vis[j] = false
}
vis[j] = true
t = append(t, cs[j])
dfs(i + 1)
t = t[:len(t)-1]
vis[j] = false
}
}
dfs(0)
@ -183,25 +174,23 @@ func permutation(S string) (ans []string) {
```ts
function permutation(S: string): string[] {
const cs: string[] = S.split('').sort();
const ans: string[] = [];
const n = cs.length;
const s: string[] = S.split('').sort();
const n = s.length;
const t = Array(n).fill('');
const vis: boolean[] = Array(n).fill(false);
const t: string[] = [];
const ans: string[] = [];
const dfs = (i: number) => {
if (i === n) {
if (i >= n) {
ans.push(t.join(''));
return;
}
for (let j = 0; j < n; ++j) {
if (vis[j] || (j > 0 && !vis[j - 1] && cs[j] === cs[j - 1])) {
continue;
if (!vis[j] && (j === 0 || s[j] !== s[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push(cs[j]);
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
dfs(0);
@ -217,25 +206,23 @@ function permutation(S: string): string[] {
* @return {string[]}
*/
var permutation = function (S) {
const cs = S.split('').sort();
const ans = [];
const n = cs.length;
const s = S.split('').sort();
const n = s.length;
const t = Array(n).fill('');
const vis = Array(n).fill(false);
const t = [];
const ans = [];
const dfs = i => {
if (i === n) {
if (i >= n) {
ans.push(t.join(''));
return;
}
for (let j = 0; j < n; ++j) {
if (vis[j] || (j > 0 && !vis[j - 1] && cs[j] === cs[j - 1])) {
continue;
if (!vis[j] && (j === 0 || s[j] !== s[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push(cs[j]);
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
dfs(0);
@ -247,37 +234,31 @@ var permutation = function (S) {
```swift
class Solution {
private var n: Int = 0
private var cs: [Character] = []
private var ans: [String] = []
private var vis: [Bool] = []
private var t: String = ""
func permutation(_ S: String) -> [String] {
cs = Array(S)
n = cs.count
cs.sort()
vis = Array(repeating: false, count: n)
var ans: [String] = []
var s: [Character] = Array(S).sorted()
var t: [Character] = Array(repeating: " ", count: s.count)
var vis: [Bool] = Array(repeating: false, count: s.count)
let n = s.count
func dfs(_ i: Int) {
if i >= n {
ans.append(String(t))
return
}
for j in 0..<n {
if !vis[j] && (j == 0 || s[j] != s[j - 1] || vis[j - 1]) {
vis[j] = true
t[i] = s[j]
dfs(i + 1)
vis[j] = false
}
}
}
dfs(0)
return ans
}
private func dfs(_ i: Int) {
if i == n {
ans.append(t)
return
}
for j in 0..<n {
if vis[j] || (j > 0 && !vis[j - 1] && cs[j] == cs[j - 1]) {
continue
}
vis[j] = true
t.append(cs[j])
dfs(i + 1)
t.removeLast()
vis[j] = false
}
}
}
```

209
lcci/08.08.Permutation II/README_EN.md
View File

@ -45,16 +45,16 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcci/08.08.Permutation%20I
### Solution 1: Sorting + Backtracking
We can first sort the string by characters, which allows us to put duplicate characters together and makes it easier for us to remove duplicates.
We can first sort the string by characters, so that duplicate characters are placed together, making it easier to remove duplicates.
Then, we design a function $dfs(i)$, which means that we need to fill in the character at the $i$-th position. The specific implementation of the function is as follows:
Then, we design a function $\textit{dfs}(i)$, which represents the character that needs to be filled at the $i$-th position. The specific implementation of the function is as follows:
- If $i = n$, it means that we have finished filling in, add the current permutation to the answer array, and then return.
- Otherwise, we enumerate the character $s[j]$ at the $i$-th position, where the range of $j$ is $[0, n - 1]$. We need to ensure that $s[j]$ has not been used and is different from the previously enumerated characters, so as to ensure that the current permutation is not repeated. If the conditions are met, we can fill in $s[j]$, and continue to recursively fill in the next position, that is, call $dfs(i + 1)$. After the recursive call ends, we need to mark $s[j]$ as unused for later enumeration.
- If $i = n$, it means we have filled all positions, add the current permutation to the answer array, and then return.
- Otherwise, we enumerate the character $\textit{s}[j]$ for the $i$-th position, where the range of $j$ is $[0, n - 1]$. We need to ensure that $\textit{s}[j]$ has not been used and is different from the previously enumerated character to ensure that the current permutation is not duplicated. If the conditions are met, we can fill $\textit{s}[j]$ and continue to recursively fill the next position by calling $\textit{dfs}(i + 1)$. After the recursive call ends, we need to mark $\textit{s}[j]$ as unused to facilitate subsequent enumeration.
In the main function, we first sort the string, then call $dfs(0)$, that is, start filling from the $0$-th position, and finally return the answer array.
In the main function, we first sort the string, then call $\textit{dfs}(0)$ to start filling from the 0th position, and finally return the answer array.
The time complexity is $O(n \times n!)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$. $n!$ enumerations need to be performed, and each enumeration requires $O(n)$ time to determine whether it is repeated. In addition, we need a marker array to mark whether each position has been used, so the space complexity is $O(n)$.
The time complexity is $O(n \times n!)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$. We need to perform $n!$ enumerations, and each enumeration requires $O(n)$ time to check for duplicates. Additionally, we need a marker array to mark whether each position has been used, so the space complexity is $O(n)$.
<!-- tabs:start -->
@ -64,21 +64,20 @@ The time complexity is $O(n \times n!)$, and the space complexity is $O(n)$. Her
class Solution:
def permutation(self, S: str) -> List[str]:
def dfs(i: int):
if i == n:
if i >= n:
ans.append("".join(t))
return
for j in range(n):
if vis[j] or (j and cs[j] == cs[j - 1] and not vis[j - 1]):
continue
t[i] = cs[j]
vis[j] = True
dfs(i + 1)
vis[j] = False
for j, c in enumerate(s):
if not vis[j] and (j == 0 or s[j] != s[j - 1] or vis[j - 1]):
vis[j] = True
t[i] = c
dfs(i + 1)
vis[j] = False
cs = sorted(S)
n = len(cs)
s = sorted(S)
ans = []
t = [None] * n
t = s[:]
n = len(s)
vis = [False] * n
dfs(0)
return ans
@ -88,35 +87,33 @@ class Solution:
```java
class Solution {
private int n;
private char[] cs;
private List<String> ans = new ArrayList<>();
private char[] s;
private char[] t;
private boolean[] vis;
private StringBuilder t = new StringBuilder();
private List<String> ans = new ArrayList<>();
public String[] permutation(String S) {
cs = S.toCharArray();
n = cs.length;
Arrays.sort(cs);
int n = S.length();
s = S.toCharArray();
Arrays.sort(s);
t = new char[n];
vis = new boolean[n];
dfs(0);
return ans.toArray(new String[0]);
}
private void dfs(int i) {
if (i == n) {
ans.add(t.toString());
if (i >= s.length) {
ans.add(new String(t));
return;
}
for (int j = 0; j < n; ++j) {
if (vis[j] || (j > 0 && !vis[j - 1] && cs[j] == cs[j - 1])) {
continue;
for (int j = 0; j < s.length; ++j) {
if (!vis[j] && (j == 0 || s[j] != s[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.append(cs[j]);
dfs(i + 1);
t.deleteCharAt(t.length() - 1);
vis[j] = false;
}
}
}
@ -128,26 +125,23 @@ class Solution {
class Solution {
public:
vector<string> permutation(string S) {
vector<char> cs(S.begin(), S.end());
sort(cs.begin(), cs.end());
int n = cs.size();
vector<string> ans;
ranges::sort(S);
string t = S;
int n = t.size();
vector<bool> vis(n);
string t;
function<void(int)> dfs = [&](int i) {
if (i == n) {
ans.push_back(t);
vector<string> ans;
auto dfs = [&](this auto&& dfs, int i) {
if (i >= n) {
ans.emplace_back(t);
return;
}
for (int j = 0; j < n; ++j) {
if (vis[j] || (j && !vis[j - 1] && cs[j] == cs[j - 1])) {
continue;
if (!vis[j] && (j == 0 || S[j] != S[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = S[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push_back(cs[j]);
dfs(i + 1);
t.pop_back();
vis[j] = false;
}
};
dfs(0);
@ -160,26 +154,23 @@ public:
```go
func permutation(S string) (ans []string) {
cs := []byte(S)
sort.Slice(cs, func(i, j int) bool { return cs[i] < cs[j] })
t := []byte{}
n := len(cs)
vis := make([]bool, n)
s := []byte(S)
sort.Slice(s, func(i, j int) bool { return s[i] < s[j] })
t := slices.Clone(s)
vis := make([]bool, len(s))
var dfs func(int)
dfs = func(i int) {
if i == n {
if i >= len(s) {
ans = append(ans, string(t))
return
}
for j := 0; j < n; j++ {
if vis[j] || (j > 0 && !vis[j-1] && cs[j] == cs[j-1]) {
continue
for j := range s {
if !vis[j] && (j == 0 || s[j] != s[j-1] || vis[j-1]) {
vis[j] = true
t[i] = s[j]
dfs(i + 1)
vis[j] = false
}
vis[j] = true
t = append(t, cs[j])
dfs(i + 1)
t = t[:len(t)-1]
vis[j] = false
}
}
dfs(0)
@ -191,25 +182,23 @@ func permutation(S string) (ans []string) {
```ts
function permutation(S: string): string[] {
const cs: string[] = S.split('').sort();
const ans: string[] = [];
const n = cs.length;
const s: string[] = S.split('').sort();
const n = s.length;
const t = Array(n).fill('');
const vis: boolean[] = Array(n).fill(false);
const t: string[] = [];
const ans: string[] = [];
const dfs = (i: number) => {
if (i === n) {
if (i >= n) {
ans.push(t.join(''));
return;
}
for (let j = 0; j < n; ++j) {
if (vis[j] || (j > 0 && !vis[j - 1] && cs[j] === cs[j - 1])) {
continue;
if (!vis[j] && (j === 0 || s[j] !== s[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push(cs[j]);
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
dfs(0);
@ -225,25 +214,23 @@ function permutation(S: string): string[] {
* @return {string[]}
*/
var permutation = function (S) {
const cs = S.split('').sort();
const ans = [];
const n = cs.length;
const s = S.split('').sort();
const n = s.length;
const t = Array(n).fill('');
const vis = Array(n).fill(false);
const t = [];
const ans = [];
const dfs = i => {
if (i === n) {
if (i >= n) {
ans.push(t.join(''));
return;
}
for (let j = 0; j < n; ++j) {
if (vis[j] || (j > 0 && !vis[j - 1] && cs[j] === cs[j - 1])) {
continue;
if (!vis[j] && (j === 0 || s[j] !== s[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push(cs[j]);
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
dfs(0);
@ -255,37 +242,31 @@ var permutation = function (S) {
```swift
class Solution {
private var n: Int = 0
private var cs: [Character] = []
private var ans: [String] = []
private var vis: [Bool] = []
private var t: String = ""
func permutation(_ S: String) -> [String] {
cs = Array(S)
n = cs.count
cs.sort()
vis = Array(repeating: false, count: n)
var ans: [String] = []
var s: [Character] = Array(S).sorted()
var t: [Character] = Array(repeating: " ", count: s.count)
var vis: [Bool] = Array(repeating: false, count: s.count)
let n = s.count
func dfs(_ i: Int) {
if i >= n {
ans.append(String(t))
return
}
for j in 0..<n {
if !vis[j] && (j == 0 || s[j] != s[j - 1] || vis[j - 1]) {
vis[j] = true
t[i] = s[j]
dfs(i + 1)
vis[j] = false
}
}
}
dfs(0)
return ans
}
private func dfs(_ i: Int) {
if i == n {
ans.append(t)
return
}
for j in 0..<n {
if vis[j] || (j > 0 && !vis[j - 1] && cs[j] == cs[j - 1]) {
continue
}
vis[j] = true
t.append(cs[j])
dfs(i + 1)
t.removeLast()
vis[j] = false
}
}
}
```

29
lcci/08.08.Permutation II/Solution.cpp
View File

@ -1,29 +1,26 @@
class Solution {
public:
vector<string> permutation(string S) {
vector<char> cs(S.begin(), S.end());
sort(cs.begin(), cs.end());
int n = cs.size();
vector<string> ans;
ranges::sort(S);
string t = S;
int n = t.size();
vector<bool> vis(n);
string t;
function<void(int)> dfs = [&](int i) {
if (i == n) {
ans.push_back(t);
vector<string> ans;
auto dfs = [&](this auto&& dfs, int i) {
if (i >= n) {
ans.emplace_back(t);
return;
}
for (int j = 0; j < n; ++j) {
if (vis[j] || (j && !vis[j - 1] && cs[j] == cs[j - 1])) {
continue;
if (!vis[j] && (j == 0 || S[j] != S[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = S[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push_back(cs[j]);
dfs(i + 1);
t.pop_back();
vis[j] = false;
}
};
dfs(0);
return ans;
}
};
};

27
lcci/08.08.Permutation II/Solution.go
View File

@ -1,26 +1,23 @@
func permutation(S string) (ans []string) {
cs := []byte(S)
sort.Slice(cs, func(i, j int) bool { return cs[i] < cs[j] })
t := []byte{}
n := len(cs)
vis := make([]bool, n)
s := []byte(S)
sort.Slice(s, func(i, j int) bool { return s[i] < s[j] })
t := slices.Clone(s)
vis := make([]bool, len(s))
var dfs func(int)
dfs = func(i int) {
if i == n {
if i >= len(s) {
ans = append(ans, string(t))
return
}
for j := 0; j < n; j++ {
if vis[j] || (j > 0 && !vis[j-1] && cs[j] == cs[j-1]) {
continue
for j := range s {
if !vis[j] && (j == 0 || s[j] != s[j-1] || vis[j-1]) {
vis[j] = true
t[i] = s[j]
dfs(i + 1)
vis[j] = false
}
vis[j] = true
t = append(t, cs[j])
dfs(i + 1)
t = t[:len(t)-1]
vis[j] = false
}
}
dfs(0)
return
}
}

34
lcci/08.08.Permutation II/Solution.java
View File

@ -1,33 +1,31 @@
class Solution {
private int n;
private char[] cs;
private List<String> ans = new ArrayList<>();
private char[] s;
private char[] t;
private boolean[] vis;
private StringBuilder t = new StringBuilder();
private List<String> ans = new ArrayList<>();
public String[] permutation(String S) {
cs = S.toCharArray();
n = cs.length;
Arrays.sort(cs);
int n = S.length();
s = S.toCharArray();
Arrays.sort(s);
t = new char[n];
vis = new boolean[n];
dfs(0);
return ans.toArray(new String[0]);
}
private void dfs(int i) {
if (i == n) {
ans.add(t.toString());
if (i >= s.length) {
ans.add(new String(t));
return;
}
for (int j = 0; j < n; ++j) {
if (vis[j] || (j > 0 && !vis[j - 1] && cs[j] == cs[j - 1])) {
continue;
for (int j = 0; j < s.length; ++j) {
if (!vis[j] && (j == 0 || s[j] != s[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.append(cs[j]);
dfs(i + 1);
t.deleteCharAt(t.length() - 1);
vis[j] = false;
}
}
}
}

22
lcci/08.08.Permutation II/Solution.js
View File

@ -3,25 +3,23 @@
* @return {string[]}
*/
var permutation = function (S) {
const cs = S.split('').sort();
const ans = [];
const n = cs.length;
const s = S.split('').sort();
const n = s.length;
const t = Array(n).fill('');
const vis = Array(n).fill(false);
const t = [];
const ans = [];
const dfs = i => {
if (i === n) {
if (i >= n) {
ans.push(t.join(''));
return;
}
for (let j = 0; j < n; ++j) {
if (vis[j] || (j > 0 && !vis[j - 1] && cs[j] === cs[j - 1])) {
continue;
if (!vis[j] && (j === 0 || s[j] !== s[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push(cs[j]);
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
dfs(0);

21
lcci/08.08.Permutation II/Solution.py
View File

@ -1,21 +1,20 @@
class Solution:
def permutation(self, S: str) -> List[str]:
def dfs(i: int):
if i == n:
if i >= n:
ans.append("".join(t))
return
for j in range(n):
if vis[j] or (j and cs[j] == cs[j - 1] and not vis[j - 1]):
continue
t[i] = cs[j]
vis[j] = True
dfs(i + 1)
vis[j] = False
for j, c in enumerate(s):
if not vis[j] and (j == 0 or s[j] != s[j - 1] or vis[j - 1]):
vis[j] = True
t[i] = c
dfs(i + 1)
vis[j] = False
cs = sorted(S)
n = len(cs)
s = sorted(S)
ans = []
t = [None] * n
t = s[:]
n = len(s)
vis = [False] * n
dfs(0)
return ans

48
lcci/08.08.Permutation II/Solution.swift
View File

@ -1,33 +1,27 @@
class Solution {
private var n: Int = 0
private var cs: [Character] = []
private var ans: [String] = []
private var vis: [Bool] = []
private var t: String = ""
func permutation(_ S: String) -> [String] {
cs = Array(S)
n = cs.count
cs.sort()
vis = Array(repeating: false, count: n)
var ans: [String] = []
var s: [Character] = Array(S).sorted()
var t: [Character] = Array(repeating: " ", count: s.count)
var vis: [Bool] = Array(repeating: false, count: s.count)
let n = s.count
func dfs(_ i: Int) {
if i >= n {
ans.append(String(t))
return
}
for j in 0..<n {
if !vis[j] && (j == 0 || s[j] != s[j - 1] || vis[j - 1]) {
vis[j] = true
t[i] = s[j]
dfs(i + 1)
vis[j] = false
}
}
}
dfs(0)
return ans
}
private func dfs(_ i: Int) {
if i == n {
ans.append(t)
return
}
for j in 0..<n {
if vis[j] || (j > 0 && !vis[j - 1] && cs[j] == cs[j - 1]) {
continue
}
vis[j] = true
t.append(cs[j])
dfs(i + 1)
t.removeLast()
vis[j] = false
}
}
}

22
lcci/08.08.Permutation II/Solution.ts
View File

@ -1,23 +1,21 @@
function permutation(S: string): string[] {
const cs: string[] = S.split('').sort();
const ans: string[] = [];
const n = cs.length;
const s: string[] = S.split('').sort();
const n = s.length;
const t = Array(n).fill('');
const vis: boolean[] = Array(n).fill(false);
const t: string[] = [];
const ans: string[] = [];
const dfs = (i: number) => {
if (i === n) {
if (i >= n) {
ans.push(t.join(''));
return;
}
for (let j = 0; j < n; ++j) {
if (vis[j] || (j > 0 && !vis[j - 1] && cs[j] === cs[j - 1])) {
continue;
if (!vis[j] && (j === 0 || s[j] !== s[j - 1] || vis[j - 1])) {
vis[j] = true;
t[i] = s[j];
dfs(i + 1);
vis[j] = false;
}
vis[j] = true;
t.push(cs[j]);
dfs(i + 1);
t.pop();
vis[j] = false;
}
};
dfs(0);

3
lcci/08.09.Bracket/README.md
View File

@ -104,8 +104,7 @@ class Solution {
public:
vector<string> generateParenthesis(int n) {
vector<string> ans;
function<void(int, int, string)> dfs;
dfs = [&](int l, int r, string t) {
auto dfs = [&](this auto&& dfs, int l, int r, string t) {
if (l > n || r > n || l < r) return;
if (l == n && r == n) {
ans.push_back(t);

3
lcci/08.09.Bracket/README_EN.md
View File

@ -112,8 +112,7 @@ class Solution {
public:
vector<string> generateParenthesis(int n) {
vector<string> ans;
function<void(int, int, string)> dfs;
dfs = [&](int l, int r, string t) {
auto dfs = [&](this auto&& dfs, int l, int r, string t) {
if (l > n || r > n || l < r) return;
if (l == n && r == n) {
ans.push_back(t);

5
lcci/08.09.Bracket/Solution.cpp
View File

@ -2,8 +2,7 @@ class Solution {
public:
vector<string> generateParenthesis(int n) {
vector<string> ans;
function<void(int, int, string)> dfs;
dfs = [&](int l, int r, string t) {
auto dfs = [&](this auto&& dfs, int l, int r, string t) {
if (l > n || r > n || l < r) return;
if (l == n && r == n) {
ans.push_back(t);
@ -15,4 +14,4 @@ public:
dfs(0, 0, "");
return ans;
}
};
};