算法leetcode｜18. 四数之和（rust重拳出击）
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文章目录
18. 四数之和：
给你一个由 n 个整数组成的数组 nums ，和一个目标值 target 。请你找出并返回满足下述全部条件且不重复的四元组 [nums[a], nums[b], nums[c], nums[d]] （若两个四元组元素一一对应，则认为两个四元组重复）：
0 <= a, b, c, d < na、b、c和d互不相同nums[a] + nums[b] + nums[c] + nums[d] == target
你可以按 任意顺序 返回答案 。
样例 1：
输入：
nums = [1,0,-1,0,-2,2], target = 0
输出：
[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
样例 2：
输入：
nums = [2,2,2,2,2], target = 8
输出：
[[2,2,2,2]]
提示：
- 1 <= nums.length <= 200
- -109 <= nums[i] <= 109
- -109 <= target <= 109
原题传送门：
https://leetcode.cn/problems/4sum/
分析
- 面对这道算法题目，二当家的陷入了沉思。
- 第一反应就是暴力四层循环，但是据说会超时。
- 如果是两数之和，在知道第一个数的情况下，第二个数也就固定了；
- 四数之和在第一个数固定后，并不能固定第二个数，第三个数和第四个数，所以才需要四层循环。
- 然而如果我们把数组先排序一下，在确定了前两个数之后，后两个数就有了关系，一个数越大，另一个数就需要越小，可以合并成一次循环。
题解
rust
impl Solution
pub fn four_sum(mut nums: Vec<i32>, target: i32) -> Vec<Vec<i32>>
let mut ans = Vec::new();
let n = nums.len();
if n < 4
return ans;
nums.sort();
for first in 0..n - 3
if first > 0 && nums[first] == nums[first - 1]
continue;
if nums[first] as i64 + nums[first + 1] as i64 + nums[first + 2] as i64 + nums[first + 3] as i64 > target as i64
break;
if target as i64 > nums[first] as i64 + nums[n - 3] as i64 + nums[n - 2] as i64 + nums[n - 1] as i64
continue;
for second in first + 1..n - 2
if second > first + 1 && nums[second] == nums[second - 1]
continue;
if nums[first] as i64 + nums[second] as i64 + nums[second + 1] as i64 + nums[second + 2] as i64 > target as i64
break;
if target as i64 > nums[first] as i64 + nums[second] as i64 + nums[n - 2] as i64 + nums[n - 1] as i64
continue;
let mut fourth = n - 1;
let t = target as i64 - nums[first] as i64 - nums[second] as i64;
for third in second + 1..n - 1
if third > second + 1 && nums[third] == nums[third - 1]
continue;
while third < fourth && nums[third] as i64 + nums[fourth] as i64 > t
fourth -= 1;
if third == fourth
break;
if nums[third] as i64 + nums[fourth] as i64 == t
ans.push(vec![nums[first], nums[second], nums[third], nums[fourth]]);
return ans;
go
func fourSum(nums []int, target int) [][]int
n := len(nums)
if n < 4
return nil
sort.Ints(nums)
var ans [][]int
for first := 0; first < n-3; first++
if first > 0 && nums[first] == nums[first-1]
continue
if nums[first]+nums[first+1]+nums[first+2]+nums[first+3] > target
break
if nums[first]+nums[n-3]+nums[n-2]+nums[n-1] < target
continue
for second := first + 1; second < n-2; second++
if second > first+1 && nums[second] == nums[second-1]
continue
if nums[first]+nums[second]+nums[second+1]+nums[second+2] > target
break
if nums[first]+nums[second]+nums[n-2]+nums[n-1] < target
continue
fourth := n - 1
t := target - nums[first] - nums[second]
for third := second + 1; third < n-1; third++
if third > second+1 && nums[third] == nums[third-1]
continue
for third < fourth && nums[third]+nums[fourth] > t
fourth -= 1
if third == fourth
break
if nums[third]+nums[fourth] == t
ans = append(ans, []intnums[first], nums[second], nums[third], nums[fourth])
return ans
c++
class Solution
public:
vector<vector<int>> fourSum(vector<int>& nums, int target)
vector<vector<int>> ans;
int n = nums.size();
if (n < 4)
return ans;
sort(nums.begin(), nums.end());
for (int first = 0; first < n - 3; ++first)
if (first > 0 && nums[first] == nums[first - 1])
continue;
if ((long) nums[first] + nums[first + 1] + nums[first + 2] + nums[first + 3] > target)
break;
if ((long) nums[first] + nums[n - 3] + nums[n - 2] + nums[n - 1] < target)
continue;
for (int second = first + 1; second < n - 2; ++second)
if (second > first + 1 && nums[second] == nums[second - 1])
continue;
if ((long) nums[first] + nums[second] + nums[second + 1] + nums[second + 2] > target)
break;
if ((long) nums[first] + nums[second] + nums[n - 2] + nums[n - 1] < target)
continue;
int fourth = n - 1;
int t = target - nums[first] - nums[second];
for (int third = second + 1; third < n - 1; ++third)
if (third > second + 1 && nums[third] == nums[third - 1])
continue;
while (third < fourth && nums[third] + nums[fourth] > t)
fourth -= 1;
if (third == fourth)
break;
if (nums[third] + nums[fourth] == t)
ans.push_back(nums[first], nums[second], nums[third], nums[fourth]);
return ans;
;
java
class Solution
public List<List<Integer>> fourSum(int[] nums, int target)
List<List<Integer>> ans = new ArrayList<>();
int n = nums.length;
if (n < 4)
return ans;
Arrays.sort(nums);
for (int first = 0; first < n - 3; ++first)
if (first > 0 && nums[first] == nums[first - 1])
continue;
if ((long) nums[first] + nums[first + 1] + nums[first + 2] + nums[first + 3] > target)
break;
if ((long) nums[first] + nums[n - 3] + nums[n - 2] + nums[n - 1] < target)
continue;
for (int second = first + 1; second < n - 2; ++second)
if (second > first + 1 && nums[second] == nums[second - 1])
continue;
if ((long) nums[first] + nums[second] + nums[second + 1] + nums[second + 2] > target)
break;
if ((long) nums[first] + nums[second] + nums[n - 2] + nums[n - 1] < target)
continue;
int fourth = n - 1;
int t = target - nums[first] - nums[second];
for (int third = second + 1; third < n - 1; ++third)
if (third > second + 1 && nums[third] == nums[third - 1])
continue;
while (third < fourth && nums[third] + nums[fourth] > t)
fourth -= 1;
if (third == fourth)
break;
if (nums[third] + nums[fourth] == t)
ans.add(Arrays.asList(nums[first], nums[second], nums[third], nums[fourth]));
return ans;
python
class Solution:
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
ans = list()
n = len(nums)
if n < 4:
return ans
nums.sort()
for first in range(n - 3):
if first > 0 and nums[first] == nums[first - 1]:
continue
if nums[first] + nums[first + 1] + nums[first + 2] + nums[first + 3] > target:
break
if nums[first] + nums[n - 3] + nums[n - 2] + nums[n - 1] < target:
continue
for second in range(first + 1, n - 2):
if second > first + 1 and nums[second] == nums[second - 1]:
continue
if nums[first] + nums[second] + nums[second + 1] + nums[second + 2] > target:
break
if nums[first] + nums[second] + nums[n - Python描述 LeetCode 18. 四数之和