523 lines
17 KiB
Markdown
523 lines
17 KiB
Markdown
<p align="center">
|
||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||
<img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20210924105952.png" width="1000"/>
|
||
</a>
|
||
<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
|
||
|
||
|
||
> 一样的道理，能解决四数之和
|
||
> 那么五数之和、六数之和、N数之和呢？
|
||
|
||
# 第18题. 四数之和
|
||
|
||
[力扣题目链接](https://leetcode-cn.com/problems/4sum/)
|
||
|
||
题意：给定一个包含 n 个整数的数组 nums 和一个目标值 target，判断 nums 中是否存在四个元素 a，b，c 和 d ，使得 a + b + c + d 的值与 target 相等？找出所有满足条件且不重复的四元组。
|
||
|
||
**注意：**
|
||
|
||
答案中不可以包含重复的四元组。
|
||
|
||
示例：
|
||
给定数组 nums = [1, 0, -1, 0, -2, 2]，和 target = 0。
|
||
满足要求的四元组集合为：
|
||
[
|
||
[-1, 0, 0, 1],
|
||
[-2, -1, 1, 2],
|
||
[-2, 0, 0, 2]
|
||
]
|
||
|
||
# 思路
|
||
|
||
四数之和，和[15.三数之和](https://programmercarl.com/0015.三数之和.html)是一个思路，都是使用双指针法, 基本解法就是在[15.三数之和](https://programmercarl.com/0015.三数之和.html) 的基础上再套一层for循环。
|
||
|
||
但是有一些细节需要注意，例如： 不要判断`nums[k] > target` 就返回了，三数之和 可以通过 `nums[i] > 0` 就返回了，因为 0 已经是确定的数了，四数之和这道题目 target是任意值。（大家亲自写代码就能感受出来）
|
||
|
||
[15.三数之和](https://programmercarl.com/0015.三数之和.html)的双指针解法是一层for循环num[i]为确定值，然后循环内有left和right下标作为双指针，找到nums[i] + nums[left] + nums[right] == 0。
|
||
|
||
四数之和的双指针解法是两层for循环nums[k] + nums[i]为确定值，依然是循环内有left和right下标作为双指针，找出nums[k] + nums[i] + nums[left] + nums[right] == target的情况，三数之和的时间复杂度是$O(n^2)$，四数之和的时间复杂度是$O(n^3)$ 。
|
||
|
||
那么一样的道理，五数之和、六数之和等等都采用这种解法。
|
||
|
||
对于[15.三数之和](https://programmercarl.com/0015.三数之和.html)双指针法就是将原本暴力$O(n^3)$的解法，降为$O(n^2)$的解法，四数之和的双指针解法就是将原本暴力$O(n^4)$的解法，降为$O(n^3)$的解法。
|
||
|
||
之前我们讲过哈希表的经典题目：[454.四数相加II](https://programmercarl.com/0454.四数相加II.html)，相对于本题简单很多，因为本题是要求在一个集合中找出四个数相加等于target，同时四元组不能重复。
|
||
|
||
而[454.四数相加II](https://programmercarl.com/0454.四数相加II.html)是四个独立的数组，只要找到A[i] + B[j] + C[k] + D[l] = 0就可以，不用考虑有重复的四个元素相加等于0的情况，所以相对于本题还是简单了不少！
|
||
|
||
我们来回顾一下，几道题目使用了双指针法。
|
||
|
||
双指针法将时间复杂度：$O(n^2)$的解法优化为 $O(n)$的解法。也就是降一个数量级，题目如下：
|
||
|
||
* [27.移除元素](https://programmercarl.com/0027.移除元素.html)
|
||
* [15.三数之和](https://programmercarl.com/0015.三数之和.html)
|
||
* [18.四数之和](https://programmercarl.com/0018.四数之和.html)
|
||
|
||
|
||
操作链表：
|
||
|
||
* [206.反转链表](https://programmercarl.com/0206.翻转链表.html)
|
||
* [19.删除链表的倒数第N个节点](https://programmercarl.com/0019.删除链表的倒数第N个节点.html)
|
||
* [面试题 02.07. 链表相交](https://programmercarl.com/面试题02.07.链表相交.html)
|
||
* [142题.环形链表II](https://programmercarl.com/0142.环形链表II.html)
|
||
|
||
双指针法在字符串题目中还有很多应用，后面还会介绍到。
|
||
|
||
C++代码
|
||
|
||
```CPP
|
||
class Solution {
|
||
public:
|
||
vector<vector<int>> fourSum(vector<int>& nums, int target) {
|
||
vector<vector<int>> result;
|
||
sort(nums.begin(), nums.end());
|
||
for (int k = 0; k < nums.size(); k++) {
|
||
// 这种剪枝是错误的，这道题目target 是任意值
|
||
// if (nums[k] > target) {
|
||
// return result;
|
||
// }
|
||
// 去重
|
||
if (k > 0 && nums[k] == nums[k - 1]) {
|
||
continue;
|
||
}
|
||
for (int i = k + 1; i < nums.size(); i++) {
|
||
// 正确去重方法
|
||
if (i > k + 1 && nums[i] == nums[i - 1]) {
|
||
continue;
|
||
}
|
||
int left = i + 1;
|
||
int right = nums.size() - 1;
|
||
while (right > left) {
|
||
// nums[k] + nums[i] + nums[left] + nums[right] > target 会溢出
|
||
if (nums[k] + nums[i] > target - (nums[left] + nums[right])) {
|
||
right--;
|
||
// 当前元素不合适了，可以去重
|
||
while (left < right && nums[right] == nums[right + 1]) right--;
|
||
// nums[k] + nums[i] + nums[left] + nums[right] < target 会溢出
|
||
} else if (nums[k] + nums[i] < target - (nums[left] + nums[right])) {
|
||
left++;
|
||
// 不合适，去重
|
||
while (left < right && nums[left] == nums[left - 1]) left++;
|
||
} else {
|
||
result.push_back(vector<int>{nums[k], nums[i], nums[left], nums[right]});
|
||
// 去重逻辑应该放在找到一个四元组之后
|
||
while (right > left && nums[right] == nums[right - 1]) right--;
|
||
while (right > left && nums[left] == nums[left + 1]) left++;
|
||
|
||
// 找到答案时，双指针同时收缩
|
||
right--;
|
||
left++;
|
||
}
|
||
}
|
||
|
||
}
|
||
}
|
||
return result;
|
||
}
|
||
};
|
||
|
||
|
||
```
|
||
|
||
|
||
|
||
|
||
|
||
## 其他语言版本
|
||
|
||
|
||
Java：
|
||
```Java
|
||
class Solution {
|
||
public List<List<Integer>> fourSum(int[] nums, int target) {
|
||
List<List<Integer>> result = new ArrayList<>();
|
||
Arrays.sort(nums);
|
||
|
||
for (int i = 0; i < nums.length; i++) {
|
||
|
||
if (i > 0 && nums[i - 1] == nums[i]) {
|
||
continue;
|
||
}
|
||
|
||
for (int j = i + 1; j < nums.length; j++) {
|
||
|
||
if (j > i + 1 && nums[j - 1] == nums[j]) {
|
||
continue;
|
||
}
|
||
|
||
int left = j + 1;
|
||
int right = nums.length - 1;
|
||
while (right > left) {
|
||
int sum = nums[i] + nums[j] + nums[left] + nums[right];
|
||
if (sum > target) {
|
||
right--;
|
||
} else if (sum < target) {
|
||
left++;
|
||
} else {
|
||
result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
|
||
|
||
while (right > left && nums[right] == nums[right - 1]) right--;
|
||
while (right > left && nums[left] == nums[left + 1]) left++;
|
||
|
||
left++;
|
||
right--;
|
||
}
|
||
}
|
||
}
|
||
}
|
||
return result;
|
||
}
|
||
}
|
||
```
|
||
|
||
Python：
|
||
```python
|
||
# 双指针法
|
||
class Solution:
|
||
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
|
||
|
||
nums.sort()
|
||
n = len(nums)
|
||
res = []
|
||
for i in range(n):
|
||
if i > 0 and nums[i] == nums[i - 1]: continue
|
||
for k in range(i+1, n):
|
||
if k > i + 1 and nums[k] == nums[k-1]: continue
|
||
p = k + 1
|
||
q = n - 1
|
||
|
||
while p < q:
|
||
if nums[i] + nums[k] + nums[p] + nums[q] > target: q -= 1
|
||
elif nums[i] + nums[k] + nums[p] + nums[q] < target: p += 1
|
||
else:
|
||
res.append([nums[i], nums[k], nums[p], nums[q]])
|
||
while p < q and nums[p] == nums[p + 1]: p += 1
|
||
while p < q and nums[q] == nums[q - 1]: q -= 1
|
||
p += 1
|
||
q -= 1
|
||
return res
|
||
```
|
||
```python
|
||
# 哈希表法
|
||
class Solution(object):
|
||
def fourSum(self, nums, target):
|
||
"""
|
||
:type nums: List[int]
|
||
:type target: int
|
||
:rtype: List[List[int]]
|
||
"""
|
||
# use a dict to store value:showtimes
|
||
hashmap = dict()
|
||
for n in nums:
|
||
if n in hashmap:
|
||
hashmap[n] += 1
|
||
else:
|
||
hashmap[n] = 1
|
||
|
||
# good thing about using python is you can use set to drop duplicates.
|
||
ans = set()
|
||
# ans = [] # save results by list()
|
||
for i in range(len(nums)):
|
||
for j in range(i + 1, len(nums)):
|
||
for k in range(j + 1, len(nums)):
|
||
val = target - (nums[i] + nums[j] + nums[k])
|
||
if val in hashmap:
|
||
# make sure no duplicates.
|
||
count = (nums[i] == val) + (nums[j] == val) + (nums[k] == val)
|
||
if hashmap[val] > count:
|
||
ans_tmp = tuple(sorted([nums[i], nums[j], nums[k], val]))
|
||
ans.add(ans_tmp)
|
||
# Avoiding duplication in list manner but it cause time complexity increases
|
||
# if ans_tmp not in ans:
|
||
# ans.append(ans_tmp)
|
||
else:
|
||
continue
|
||
return list(ans)
|
||
# if used list() to save results, just
|
||
# return ans
|
||
|
||
```
|
||
|
||
Go：
|
||
```go
|
||
func fourSum(nums []int, target int) [][]int {
|
||
if len(nums) < 4 {
|
||
return nil
|
||
}
|
||
sort.Ints(nums)
|
||
var res [][]int
|
||
for i := 0; i < len(nums)-3; i++ {
|
||
n1 := nums[i]
|
||
// if n1 > target { // 不能这样写,因为可能是负数
|
||
// break
|
||
// }
|
||
if i > 0 && n1 == nums[i-1] {
|
||
continue
|
||
}
|
||
for j := i + 1; j < len(nums)-2; j++ {
|
||
n2 := nums[j]
|
||
if j > i+1 && n2 == nums[j-1] {
|
||
continue
|
||
}
|
||
l := j + 1
|
||
r := len(nums) - 1
|
||
for l < r {
|
||
n3 := nums[l]
|
||
n4 := nums[r]
|
||
sum := n1 + n2 + n3 + n4
|
||
if sum < target {
|
||
l++
|
||
} else if sum > target {
|
||
r--
|
||
} else {
|
||
res = append(res, []int{n1, n2, n3, n4})
|
||
for l < r && n3 == nums[l+1] { // 去重
|
||
l++
|
||
}
|
||
for l < r && n4 == nums[r-1] { // 去重
|
||
r--
|
||
}
|
||
// 找到答案时,双指针同时靠近
|
||
r--
|
||
l++
|
||
}
|
||
}
|
||
}
|
||
}
|
||
return res
|
||
}
|
||
```
|
||
|
||
javaScript:
|
||
|
||
```js
|
||
/**
|
||
* @param {number[]} nums
|
||
* @param {number} target
|
||
* @return {number[][]}
|
||
*/
|
||
var fourSum = function(nums, target) {
|
||
const len = nums.length;
|
||
if(len < 4) return [];
|
||
nums.sort((a, b) => a - b);
|
||
const res = [];
|
||
for(let i = 0; i < len - 3; i++) {
|
||
// 去重i
|
||
if(i > 0 && nums[i] === nums[i - 1]) continue;
|
||
for(let j = i + 1; j < len - 2; j++) {
|
||
// 去重j
|
||
if(j > i + 1 && nums[j] === nums[j - 1]) continue;
|
||
let l = j + 1, r = len - 1;
|
||
while(l < r) {
|
||
const sum = nums[i] + nums[j] + nums[l] + nums[r];
|
||
if(sum < target) { l++; continue}
|
||
if(sum > target) { r--; continue}
|
||
res.push([nums[i], nums[j], nums[l], nums[r]]);
|
||
while(l < r && nums[l] === nums[++l]);
|
||
while(l < r && nums[r] === nums[--r]);
|
||
}
|
||
}
|
||
}
|
||
return res;
|
||
};
|
||
```
|
||
|
||
TypeScript：
|
||
|
||
```typescript
|
||
function fourSum(nums: number[], target: number): number[][] {
|
||
nums.sort((a, b) => a - b);
|
||
let first: number = 0,
|
||
second: number,
|
||
third: number,
|
||
fourth: number;
|
||
let length: number = nums.length;
|
||
let resArr: number[][] = [];
|
||
for (; first < length; first++) {
|
||
if (first > 0 && nums[first] === nums[first - 1]) {
|
||
continue;
|
||
}
|
||
for (second = first + 1; second < length; second++) {
|
||
if ((second - first) > 1 && nums[second] === nums[second - 1]) {
|
||
continue;
|
||
}
|
||
third = second + 1;
|
||
fourth = length - 1;
|
||
while (third < fourth) {
|
||
let total: number = nums[first] + nums[second] + nums[third] + nums[fourth];
|
||
if (total === target) {
|
||
resArr.push([nums[first], nums[second], nums[third], nums[fourth]]);
|
||
third++;
|
||
fourth--;
|
||
while (nums[third] === nums[third - 1]) third++;
|
||
while (nums[fourth] === nums[fourth + 1]) fourth--;
|
||
} else if (total < target) {
|
||
third++;
|
||
} else {
|
||
fourth--;
|
||
}
|
||
}
|
||
}
|
||
}
|
||
return resArr;
|
||
};
|
||
```
|
||
|
||
PHP:
|
||
|
||
```php
|
||
class Solution {
|
||
/**
|
||
* @param Integer[] $nums
|
||
* @param Integer $target
|
||
* @return Integer[][]
|
||
*/
|
||
function fourSum($nums, $target) {
|
||
$res = [];
|
||
sort($nums);
|
||
for ($i = 0; $i < count($nums); $i++) {
|
||
if ($i > 0 && $nums[$i] == $nums[$i - 1]) {
|
||
continue;
|
||
}
|
||
for ($j = $i + 1; $j < count($nums); $j++) {
|
||
if ($j > $i + 1 && $nums[$j] == $nums[$j - 1]) {
|
||
continue;
|
||
}
|
||
$left = $j + 1;
|
||
$right = count($nums) - 1;
|
||
while ($left < $right) {
|
||
$sum = $nums[$i] + $nums[$j] + $nums[$left] + $nums[$right];
|
||
if ($sum < $target) {
|
||
$left++;
|
||
}
|
||
else if ($sum > $target) {
|
||
$right--;
|
||
}
|
||
else {
|
||
$res[] = [$nums[$i], $nums[$j], $nums[$left], $nums[$right]];
|
||
while ($left < $right && $nums[$left] == $nums[$left+1]) $left++;
|
||
while ($left < $right && $nums[$right] == $nums[$right-1]) $right--;
|
||
$left++;
|
||
$right--;
|
||
}
|
||
}
|
||
}
|
||
}
|
||
return $res;
|
||
}
|
||
}
|
||
```
|
||
|
||
Swift:
|
||
```swift
|
||
func fourSum(_ nums: [Int], _ target: Int) -> [[Int]] {
|
||
var res = [[Int]]()
|
||
var sorted = nums
|
||
sorted.sort()
|
||
for k in 0 ..< sorted.count {
|
||
// 这种剪枝不行,target可能是负数
|
||
// if sorted[k] > target {
|
||
// return res
|
||
// }
|
||
// 去重
|
||
if k > 0 && sorted[k] == sorted[k - 1] {
|
||
continue
|
||
}
|
||
|
||
let target2 = target - sorted[k]
|
||
for i in (k + 1) ..< sorted.count {
|
||
if i > (k + 1) && sorted[i] == sorted[i - 1] {
|
||
continue
|
||
}
|
||
var left = i + 1
|
||
var right = sorted.count - 1
|
||
while left < right {
|
||
let sum = sorted[i] + sorted[left] + sorted[right]
|
||
if sum < target2 {
|
||
left += 1
|
||
} else if sum > target2 {
|
||
right -= 1
|
||
} else {
|
||
res.append([sorted[k], sorted[i], sorted[left], sorted[right]])
|
||
while left < right && sorted[left] == sorted[left + 1] {
|
||
left += 1
|
||
}
|
||
while left < right && sorted[right] == sorted[right - 1] {
|
||
right -= 1
|
||
}
|
||
// 找到答案 双指针同时收缩
|
||
left += 1
|
||
right -= 1
|
||
}
|
||
}
|
||
}
|
||
}
|
||
return res
|
||
}
|
||
```
|
||
|
||
C#:
|
||
```csharp
|
||
public class Solution
|
||
{
|
||
public IList<IList<int>> FourSum(int[] nums, int target)
|
||
{
|
||
var result = new List<IList<int>>();
|
||
|
||
Array.Sort(nums);
|
||
|
||
for (int i = 0; i < nums.Length - 3; i++)
|
||
{
|
||
int n1 = nums[i];
|
||
if (i > 0 && n1 == nums[i - 1])
|
||
continue;
|
||
|
||
for (int j = i + 1; j < nums.Length - 2; j++)
|
||
{
|
||
int n2 = nums[j];
|
||
if (j > i + 1 && n2 == nums[j - 1])
|
||
continue;
|
||
|
||
int left = j + 1;
|
||
int right = nums.Length - 1;
|
||
|
||
while (left < right)
|
||
{
|
||
int n3 = nums[left];
|
||
int n4 = nums[right];
|
||
int sum = n1 + n2 + n3 + n4;
|
||
|
||
if (sum > target)
|
||
{
|
||
right--;
|
||
}
|
||
else if (sum < target)
|
||
{
|
||
left++;
|
||
}
|
||
else
|
||
{
|
||
result.Add(new List<int> { n1, n2, n3, n4 });
|
||
|
||
while (left < right && nums[left] == n3)
|
||
{
|
||
left++;
|
||
}
|
||
|
||
while (left < right && nums[right] == n4)
|
||
{
|
||
right--;
|
||
}
|
||
}
|
||
}
|
||
}
|
||
}
|
||
|
||
return result;
|
||
}
|
||
}
|
||
```
|
||
|
||
-----------------------
|
||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|