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]]
Example 2:
Input: nums = []
Output: []
Example 3:
Input: nums = [0]
Output: []
Constraints:
0 <= nums.length <= 3000
-105 <= nums[i] <= 105
/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function (nums) {
nums.sort((a, b) => a - b)
const len = nums.length
const result = []
let i = 0;
while (i < len - 2) {
let l = i + 1
let r = len - 1
while (r > l) {
const sum = nums[i] + nums[l] + nums[r]
if (sum < 0) {
l++
} else if (sum > 0) {
r--
} else {
result.push([nums[i], nums[l], nums[r]])
while (l < r && nums[l + 1] === nums[l]) {
l++
}
while (r > l && nums[r - 1] === nums[r]) {
r--
}
l++
r--
}
}
while (i < len - 1 && nums[i + 1] === nums[i]) {
i++
}
i++
}
return result
}
export { threeSum }