题目描述：

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.

For example, given array S = {-1 0 1 2 -1 -4},    A solution set is:    (-1, 0, 1)    (-1, -1, 2)

 

　　读完题目，立刻想到之前做过的Two Sum，原理相同。只是我提交的时候遇到OLE(Output Limit Exceeded)，原因是输出的结果中包含重复的 triplet。

solution：

vector
     
      
        > threeSum(vector
       
         &num) {    int n = num.size();    vector
        
         
           > res;    if(n < 3)        return res;    sort(num.begin(),num.end());    vector
          
            temp(3,0); int low, high; for (int i = 0;i < n-2;++i) { low = i + 1; high = n - 1; while (low < high) { if(num[i] + num[low] + num[high] == 0) { temp[0] = num[i]; temp[1] = num[low]; temp[2] = num[high]; res.push_back(temp); ++low; --high; while (low < high && num[low] == num[low-1]) ++low; while (low < high && num[high] == num[high+1]) --high; } else if (num[i] + num[low] + num[high] < 0) ++low; else --high; } while (i+1 < n-2 && num[i+1] == num[i]) ++i; } return res;}
          
         
        
       
      
     

　　想要AC，一是思路，二是细节。Fighting!!!