LeetCode in TypeScript
15. 3Sum
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
Solution
function threeSum(nums: number[]): number[][] { //NOSONAR
nums.sort((a, b) => a - b)
const len = nums.length
const result: number[][] = []
let l: number
let r: number
for (let i = 0; i < len - 2; i++) {
l = i + 1
r = len - 1
while (r > l) {
const sum = nums[i] + nums[l] + nums[r]
if (sum < 0) {
l++
} else if (sum > 0) {
r--
} else {
const list: number[] = [nums[i], nums[l], nums[r]]
result.push(list)
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++ //NOSONAR
}
}
return result
}
export { threeSum }