foursum 问题

问题描述:

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)

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int i , j , m , n;
        set<vector<int> >revalue;
        vector<int> entry(4);
        int count=num.size();
        sort(num.begin(),num.end());
        for(i=0 ; i< count ; i++){
            int middle=target-num[i];//num combination
            for(j=i+1 ; j< count ; j++){
                m=j+1;
                n=count-1;
                int inner=middle-num[j];
                for(;m<n;){
                    if(num[m]+num[n]<inner)
                        m++;
                    else if(num[m]+num[n]==inner){
                        entry[0]=num[i];
                        entry[1]=num[j];
                        entry[2]=num[m];
                        entry[3]=num[n];
                        sort(entry.begin(),entry.end());
                        revalue.insert(entry);
                        m++;
                        n--;
                    }
                    else
                        n--;
                }
            }
        }
        vector<vector<int> > last(revalue.begin(),revalue.end());
        return last;
    }
};

这是leetcode里面的foursum问题，我的解法是在threesum的基础上继续加一重循环，没想到这样居然通过了……我后来查了一下看看有没有比我的o(n3)时间复杂度还好的，暂时没有看到