#Leetcode# 18. 4Sum

https://leetcode.com/problems/4sum/

 

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

The solution set must not contain duplicate quadruplets.

Example:

Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

代码：

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        set<vector<int>> ans;
        int n = nums.size();
        for(int i = 0; i < n - 3; i ++) {
            for(int j = i + 1; j < n - 2; j ++) {
                int num = nums[i] + nums[j];
                int find = target - num;
                int l = j + 1, r = n - 1;
                while(l < r) {
                    if(nums[l] + nums[r] == find) {
                        ans.insert({nums[i], nums[j], nums[l], nums[r]});
                        while(l < r && nums[l] == nums[l + 1]) l ++;
                        while(l < r && nums[r] == nums[r - 1]) r --;
                    
                        l ++;
                        r --;
                    }
                    else if(nums[l] + nums[r] > find) r --;
                    else l ++;
                }
            }
        }
        return vector<vector<int>>(ans.begin(), ans.end());
    }
};

这个和之前的 3Sum 还有 3Sum Closest 一样的 时间复杂度是 $O(n ^ 3)$ 不知道还有没有更简单的方法 

emmmm 这个颜色好像芋泥 嘻嘻