LeetCode | 18. 4Sum

Given an array S of n integers, are there elements a, b, c, and d in S 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.

For example, given array S = [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]]

跟2Sum 3Sum类似,用两个指针left , right 在后面遍历一遍,复杂度 O(n^3)
55ms Ac

class Solution {public:        vector<vector<int>> fourSum(vector<int>& nums, int target)        {            vector<vector<int> > res;            vector<int> tmp;            sort(nums.begin(),nums.end());            int len = nums.size();            for(int i=0;i<len;i++)            {                for(int j=i+1;j<len;j++)                {                    int left = j+1, right = len-1;                    while(left < right)                    {                        int sum = nums[i]+nums[j]+nums[left]+nums[right];                        if(sum == target)//找到目标                        {                            tmp.push_back(nums[i]); tmp.push_back(nums[j]);                            tmp.push_back(nums[left]); tmp.push_back(nums[right]);                            res.push_back(tmp);                            while(left<len-1 && nums[left]==tmp[2])//不这样比较容易陷入死循环:因为left/right可能会没有移动                            {                                left++;                            }                            while(right>=1 && nums[right]==tmp[3])                            {                                right--;                            }                            tmp.clear();                        }                        else if(sum < target)                        {                            left++;                        }                        else                        {                            right--;                        }                        while(j+1<len && nums[j]==nums[j+1])                        {                            j++;                        }                    }                }                while(i+1<len && nums[i]==nums[i+1])                {                    i++;                }            }            return res;        }};