3Sum

Problem Description

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.

Constraints:

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

Solution (JavaScript)

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function(nums) {
	var rtn = [];
	if (nums.length < 3) {
		return rtn;
	}
	nums = nums.sort(function(a, b) {
		return a - b;
	});
	for (var i = 0; i < nums.length - 2; i++) {
		if (nums[i] > 0) {
			return rtn;
		}
		if (i > 0 && nums[i] == nums[i - 1]) {
			continue;
		}
		for (var j = i + 1, k = nums.length - 1; j < k;) {
			if (nums[i] + nums[j] + nums[k] === 0) {
				rtn.push([nums[i], nums[j], nums[k]]);
				j++;
				k--;
				while (j < k && nums[j] == nums[j - 1]) {
					j++;
				}
				while (j < k && nums[k] == nums[k + 1]) {
					k--;
				}
			} else if (nums[i] + nums[j] + nums[k] > 0) {
				k--;
			} else {
				j++;
			}
		}
	}
	return rtn;
};