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)

思路：首先转换为求2个数字和的情况，枚举每一位，那么这位固定后，就能根据双指针法来处理这个问题啦

 关键是去重的部分：1.  双指针法的头指针从枚举的下一位开始，这样就能排除前面枚举的结果

     2. 对于每次枚举相邻两位是一样的话，我们就跳过

     3. 对于双指针相邻两位一样的话，我们也跳过

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int>> ans;
        sort(num.begin(), num.end());

        for (int i = 0; i < num.size(); i++) {
            if (i && num[i] == num[i-1]) continue;

            int s = i + 1;
            int e = num.size() - 1;
            while (s < e) {
                if (s > i+1 && num[s] == num[s-1]) {
                    s++;
                    continue;
                } 
                if (e < num.size() - 1 && num[e] == num[e+1]) {
                    e--;
                    continue;
                }

                int sum = num[i] + num[s] + num[e];
                if (sum > 0) 
                    e--;
                else if (sum < 0)
                    s++; 
                else {
                    vector<int> ve;
                    ve.push_back(num[i]);     
                    ve.push_back(num[s]);
                    ve.push_back(num[e]);
                    ans.push_back(ve);
                    s++;
                }
            }
        }

        return ans;
    }
};

LeetCode 3Sum