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=,cnt=;
         while(NumCopy[m]<)
             ++m;
         int f = m,j=m;
         while(NumCopy[f] == ){
             ++f;
             ++cnt;
         }
         if(cnt>=){
             f = m+cnt-;
             j = m+cnt-;
         }
         for(int i=;f<NumCopy.size();++f){
             int tempi =i,tempj =j;
             int sum = - * (NumCopy[i]+NumCopy[j]);
             while( sum > NumCopy[f]&& i<j ){
                 ++i;
                 sum = -* (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 = - * (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::>_<:: 。