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)

pro:给一个数组，和一个target，求出数组中所有和等于target的元组，不能有重复的。

sol:仍旧采用了2sum的思想，不知道有没有其他好的方法。

1.对数组排序

2.用i,j枚举数组中的两个元素(一般我们说枚举就是指比如j从i+1开始)，然后m和n分别从j+1和len-1开始

code:

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