Leetcode题解-18. 4Sum

Leetcode题解-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]
]

思路

把4Sum问题化成3Sum，3Sum化成2Sum，所以还是在2Sum的基础上加两重循环。2Sum的思路是：把数组nums从小到大排序，一个指针front指向数组头，一个指针back指向数组尾，如果front和back指向数组元素之和小于target，front++，如果大于target则back–，直到front不小于back结束

代码

vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> res;        int l = nums.size();        if(l < 4) return res;        sort(nums.begin(), nums.end());        for(int i = 0; i < l-3; i++){            int ttarget = target - nums[i];            for(int j = i+1; j < l-2; j++){                int front = j + 1, back = l - 1, tttarget = ttarget - nums[j];                while(front < back){                    int sum = nums[front] + nums[back];                    if(sum < tttarget) front++;                    else if(sum > tttarget) back--;                    else{                        vector<int> tem(4,0);                        tem[0] = nums[i];                        tem[1] = nums[j];                        tem[2] = nums[front];                        tem[3] = nums[back];                        res.push_back(tem);                        while(front < back && nums[front] == tem[2]) front++;                        while(front < back && nums[back] == tem[3]) back--;                    }                }                while(nums[j] == nums[j+1]) j++;            }            while(nums[i] == nums[i+1]) i++;        }        return res;    }