LeetCode 4Sum C++

#include <iostream>
#include <vector>
#include<algorithm>
#include <queue>
using namespace std;

/************************************************************************/
/*
Problem:
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]
]

Author  :   crazys_popcorn@126.com

DateTime:   2017年8月5日  12:06:17

*/
/************************************************************************/


class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target)
    {
        vector< vector  <int > >_result;
        int _size = nums.size();
        if (nums.size() < 4)
            return _result;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < _size - 3; i++)
        {
            if (i>0 && nums[i]== nums[i-1])
                continue;
            if (nums[i]+nums[i+1]+nums[i+2] +nums[i+3]> target)
                break;
            if (nums[i] + nums[_size - 1] + nums[_size - 2] + nums[_size - 3] < target)
                continue;
            for (int j = i + 1;j <_size - 2; j++)
            {
                //为什么要用j > i+1 因为这是 要往后推迟一个。不要让他和i相比 也就是前一位数字
                if ( j > i +1 &&nums[j] == nums[j - 1])
                    continue;
                if (nums[i] + nums[j] + nums[j+1] + nums[j+2]> target)
                    break;
                if (nums[i] + nums[_size -  1] + nums[_size - 2] + nums[j] < target)
                    continue;
                int k = j + 1, m = _size - 1;
                while (k < m)
                {
                    int sum = nums[i] + nums[j] + nums[k] + nums[m];
                    if (sum <target)k++;
                    else if (sum > target) m--;
                    else
                    {
                        _result.push_back(vector<int>{ nums[i], nums[j], nums[k], nums[m] });
                        do{ k++;} while (nums[k] == nums[k - 1] && k < m);
                        do{ m--;} while (nums[m] == nums[m + 1] && k < m);
                    }
                }

            }

        }
        return _result;
    }
};

void main()
{
    Solution s1;
    //                   0     1   2   3  4  5  6   7
    vector<int> _temp = { -5, -5, -3, -1, 0, 2, 4, 5 };
    vector<vector<int>> num = s1.fourSum(_temp, -11);
    std::cout << "" << endl;
    system("pause");

}