(LeetCode 18) 4Sum

Q:
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)

题意大致是，先给你一个整型数组，让你计算出里面所有相加等于0的四个数的组合，按照升序存放。

solution：
这道题和(LeetCode 16) 3Sum Closest， (LeetCode 16) 3Sum Closest基本一致。
连剪枝方法也是一样的，需要说明的是，这道题中剪枝十分重要，第一个简单的剪枝就可以将时间效率提高一倍以上，后面几个剪枝可以吧时间消耗压缩为原来的1/5.很神奇…..

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