[leetcode] 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：我们可以仿照3sum的解决方法。这里枚举第一个和第二个数，然后对余下数的求2sum，算法复杂度为O（n^3）

参考链接：http://www.cnblogs.com/TenosDoIt/p/3649607.html

代码：

class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target) {        int n=num.size();        vector<vector<int> > res;        sort(num.begin(),num.end());        for(int i=0;i<n-3;i++){            if(i>0 && num[i]==num[i-1]) continue;            for(int j=i+1;j<n-2;j++){                if(j>i+1 && num[j]==num[j-1]) continue;                int target2=target-num[i]-num[j];                int left=j+1,right=n-1;                while(left<right){                    int temp=num[left]+num[right];                    if(temp>target2) right--;                    if(temp<target2) left++;                    if(temp==target2){                        vector<int> tempres;                        tempres.push_back(num[i]);                        tempres.push_back(num[j]);                        tempres.push_back(num[left]);                        tempres.push_back(num[right]);                        res.push_back(tempres);                        int k=left+1;                        while(k<right && num[k]==num[left]) k++;                        left=k;                        k=right-1;                        while(k>left && num[k]==num[right]) k--;                        right=k;                    }                }            }        }        return res;    }};