15 - 3 Sum

#medium

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] ]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.

Example 2:
Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.

Example 3:
Input: nums = [0,0,0]
Output: [ [0,0,0] ]
Explanation: The only possible triplet sums up to 0.

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> result;
        sort( nums.begin(), nums.end() );
        // 找出a + b + c = 0
        // a = nums[i], b = nums[left], c = nums[right]
        for ( int i = 0; i < nums.size(); i++ ) {
            if ( nums[i] > 0 ) return result;
            if ( i > 0 && nums[i] == nums[i-1] ) continue;
            int left = i + 1;
            int right = nums.size() - 1;
            while ( right > left ) {
                if ( nums[i] + nums[left] + nums[right] > 0 ) right--;
                else if ( nums[i] + nums[left] + nums[right] < 0 ) left++;
                else {
                    result.push_back( {nums[i], nums[left], nums[right] });
                    // 去重邏輯應該放在找到一個三元組之後，對b和c去重
                    while ( right > left && nums[right] == nums[right-1] ) right--;
                    while ( right > left && nums[left] == nums[left+1] ) left++;
                    right--;
                    left++;
                }

            }
        }
        return result;
    }
};