【算法】Three Sum 问题

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>& nums) {

        vector<vector<int>> results;

        sort(nums.begin(), nums.end());

        int size = nums.size();
        for(int i = 0; i < size - 2; ++i) {
            if ( i > 0 && (nums[i] == nums[i-1]) ) continue;

            int j = i + 1, k = size - 1;
            while( j < k) {
                int a = nums[i], b = nums[j], c = nums[k];

                int sum = a + b + c;
                if (sum == 0) {
                    vector<int> result {a, b, c};
                    results.push_back(result);

                    ++j;
                    --k;
                }
                else if(sum < 0){
                    ++j;
                }
                else {
                    --k;
                }

                while ( nums[j] == nums[j - 1] && j < k) {
                    ++j;
                }
                while ( nums[k] == nums[k + 1] && j < k) {
                    --k;
                }
            }
        }

        return results;
    }
};