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<vector<int> > threeSum(vector<int> &num) {
    int n = num.size();
    vector<vector<int> > res;
    if(n < 3)
        return res;
    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);
                ++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!!!

转载于:https://www.cnblogs.com/gattaca/p/4281581.html