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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:
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<vector<int> > threeSum(vector<int> &num) { int n = num.size(); vector<vector<int> > res; if(n < 3) return res; std::sort(num.begin(),num.end()); vector<int> 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); while (low < high && num[low+1] == num[low]) ++low; while (low < high && num[high-1] == num[high]) --high; ++low; --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!!!
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原文地址：http://www.cnblogs.com/gattaca/p/4281581.html
