[Leetcode] 3Sum

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)

去重比较麻烦，在外层循环同样的元素只用一次，在内层循环，每次匹配后，跳过所有与当前匹配相同的元素。

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