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:

• 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)
思路：找出四个元素，其和为制定元素的值，首先定义一个函数，来找出所有符合这一条件的方法。首先定义前两个相邻的索引，然后在定义一个这两个元素其后的头指针begin，最后定义一个尾指针end。如果target小于这四个数之和，则begin++；如果大于这四个数之和，则end--；如果相等呢，则将这四个数放入path数组中，最后放入allpaths，最后下一循环begin++；主函数里就可以围绕相邻两个元素的索引变换了，最后将这个allpaths中重复的元素去掉。

class Solution {
public:
    void findFourSum(vector<int> &num,int target,int first,int second,vector<vector<int> > &allpaths)
    {
        target=target-num[first]-num[second];
        int begin=second+1;
        int end=num.size()-1;
        while(begin<end)
        {
            int sum=num[begin]+num[end];
            if(sum>target)
                end--;
            else if(sum<target)
                begin++;
            else
            {
                vector<int> path(4);
                path[0]=num[first];
                path[1]=num[second];
                path[2]=num[begin];
                path[3]=num[end];
                allpaths.push_back(path);
                begin++;
            }
        }
    }
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        vector<vector<int> > result;
        result.clear();
        if(num.size()<4)
            return result;
        sort(num.begin(),num.end());
        vector<vector<int> > allpaths;
        allpaths.clear();
        for(int i=0;i<num.size()-3;i++)
        {
            for(int j=i+1;j<num.size()-2;j++)
            {
                findFourSum(num,target,i,j,allpaths);
            }
        }
        for(int i=0;i<allpaths.size();i++)
        {
            if(find(result.begin(),result.end(),allpaths[i])==result.end())
                result.push_back(allpaths[i]);
        }
        return result;
    }
};