4Sum

最近较忙，没怎么写LeetCode以后尽量保证每天一道题目以上。
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:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
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)

最近忙别的事情，看不懂代码的直接邮箱联系[e-mail].

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> rsvv;        if (nums.size()<4)            return rsvv;        vector<int> temp(4, 0);        sort(nums.begin(), nums.end());        int size = nums.size();        for (int i = 0; i<size - 3; i++)        for (int j = i + 1; j<size- 2; j++)        {            int m = j + 1, n = size - 1;            while (m<n)            {                int sum = nums[i] + nums[j] + nums[m] + nums[n];                if (sum<target)                    m++;                else if (sum>target)                    n--;                else{                    temp[0] = nums[i];                    temp[1] = nums[j];                    temp[2] = nums[m];                    temp[3] = nums[n];                    //if (!(rsvv.size()>0 && rsvv[rsvv.size() - 1][0] == temp[0] && rsvv[rsvv.size() - 1][1] == temp[1]                        //&& rsvv[rsvv.size() - 1][2] == temp[2] && rsvv[rsvv.size() - 1][3] == temp[3]))                        //rsvv.push_back(temp);                    if (!((j > i + 1 && nums[j] == nums[j - 1]) || (m > j + 1 && nums[m] == nums[m - 1]) || i > 0 && nums[i] == nums[i - 1]))                    {                        rsvv.push_back(temp);                    }                    m++;                }            }        }        return rsvv;    }};