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.

 1 class Solution {
 2 public:
 3     vector<vector<int> > threeSum(vector<int> &num) {
 4         vector<int> NumCopy(num);
 5         vector<vector<int> > returnvct;
 6         sort(NumCopy.begin(),NumCopy.end());
 7         int m=0,cnt=0;
 8         while(NumCopy[m]<0)
 9             ++m;
10         int f = m,j=m;
11         while(NumCopy[f] == 0){
12             ++f;
13             ++cnt;
14         }
15         if(cnt>=3){
16             f = m+cnt-1;
17             j = m+cnt-2;
18         }
19         for(int i=0;f<NumCopy.size();++f){
20             int tempi =i,tempj =j;
21             int sum = -1 * (NumCopy[i]+NumCopy[j]);
22             while( sum > NumCopy[f]&& i<j ){
23                 ++i;
24                 sum = -1* (NumCopy[i]+ NumCopy[j]);
25             }
26             if(sum == NumCopy[f]){
27                 vector<int> tmp;
28                 tmp.push_back(i);
29                 tmp.push_back(j);
30                 tmp.push_back(f);
31                 returnvct.push_back(tmp);
32             }
33             
34             while(sum < NumCopy[f] && i<j){
35                 --j;
36                 sum = -1 * (NumCopy[i]+ NumCopy[j]);
37             }
38             i = tempi;
39             j= tempj;
40         }
41         return returnvct;
42     }
43 };

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::>_<:: 。