[LeetCode] 3Sum

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

 

Example 1:

Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]]

Example 2:

Input: nums = [] Output: []

Example 3:

Input: nums = [0] Output: []

 

Constraints:

• 0 <= nums.length <= 3000
• -105 <= nums[i] <= 105

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> ans;
        sort(nums.begin(), nums.end());
        int target, left, right, sum;
        for (int i = 0; i < nums.size(); i++)
        {
            if (i > 0 && nums[i - 1] == nums[i]) continue;
            
            target = -nums[i];
            left = i + 1;
            right = nums.size() - 1;
            while (left < right)
            {
                sum = nums[left] + nums[right];
                if (sum < target) left++;
                else if (sum > target) right--;
                else
                {
                    ans.push_back({ nums[i], nums[left], nums[right] });
                    while (left < right && nums[left] == ans[ans.size() - 1][1]) left++;
                    while (left < right && nums[right] == ans[ans.size() - 1][2]) right--;
                }
            }
        }
        return ans;
    }
};