three Sum

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

• Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
• The solution set must not contain duplicate triplets.

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<int> NumCopy(num);
        vector<vector<int> > returnvct;
        sort(NumCopy.begin(),NumCopy.end());
        int m=0,cnt=0;
        while(NumCopy[m]<0)
            ++m;
        int f = m,j=m;
        while(NumCopy[f] == 0){
            ++f;
            ++cnt;
        }
        if(cnt>=3){
            f = m+cnt-1;
            j = m+cnt-2;
        }
        for(int i=0;f<NumCopy.size();++f){
            int tempi =i,tempj =j;
            int sum = -1 * (NumCopy[i]+NumCopy[j]);
            while( sum > NumCopy[f]&& i<j ){
                ++i;
                sum = -1* (NumCopy[i]+ NumCopy[j]);
            }
            if(sum == NumCopy[f]){
                vector<int> tmp;
                tmp.push_back(i);
                tmp.push_back(j);
                tmp.push_back(f);
                returnvct.push_back(tmp);
            }
            
            while(sum < NumCopy[f] && i<j){
                --j;
                sum = -1 * (NumCopy[i]+ NumCopy[j]);
            }
            i = tempi;
            j= tempj;
        }
        return returnvct;
    }
};

利用昨天twoSum的方法 将a+b+c =0 变成 a+b =-c;数组 以0为界限 分为前后两部分。

可是又超时了::>_<:: 。