Problem

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

Solution

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

Explain:

先排序，用双指针求解，注意重复的数字。

Complexity:

• Time complexity : O(n^2).
• Space complexity : O(1).