[leetcode 15] 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)

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > res;
        if (num.size() < 3) {
            return res;
        }
        const int target = 0; //可扩展为sum为任意数
        sort(num.begin(), num.end());
        auto last = num.end();
        for (auto i = num.begin(); i < last - 2; i++) {
            auto j = i + 1;
            if (i > num.begin() && *i == *(i-1)) {
                continue;
            }
            auto k = last - 1;
            while (j < k) {
                if (*i + *j + *k < 0) {
                    j++;
                    while (*j == *(j-1) && j < k) j++;
                } else if (*i + *j + *k > 0) {
                    k--;
                    while (*k == *(k+1) && j < k) k--;
                } else {
                    res.push_back({*i,*j, *k});
                    j++;
                    k--;
                    while (*j == *(j-1) && j < k) {
                        j++;
                    }
                    while (*k == *(k+1) && j < k) {
                        k--;
                    }
                }
            }
        }
        return res;
    }
};