LeetCode: 3Sum

Title: 

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)

//68ms
class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        std::sort(num.begin(), num.end());
        int target1 = 0;
        vector<int> every_vec;
        vector<vector<int> > final_vec;
        if (num.size() <= 2) return final_vec; //corner case:
        for (int i = 0; i < num.size() - 2; ++i) {
            if(i > 0 && num[i] == num[i - 1]) 
                continue;
            target1 = 0 - num[i];
            int first = i + 1;
            int last = num.size() - 1;
            while (first < last) {
                if (num[first] + num[last] < target1) {
                    while (first < num.size() - 1 && num[first] == num[first + 1]) first++;
                    first++;
                }else if (num[first] + num[last] > target1) {
                    while (last > 0 && num[last] == num[last - 1]) last--;
                    last--;
                }else {
                    every_vec.clear();
                    every_vec.push_back(num[i]);
                    every_vec.push_back(num[first]);
                    every_vec.push_back(num[last]);
                    final_vec.push_back(every_vec);
                    while (first < num.size() - 1 && num[first] == num[first+1]) first++;
                    while (last > 0 && num[last] == num[last - 1]) last--;
                    first++;
                    last--;
                }
            }
        }
        return final_vec;
    }
};