LeetCode-15. 3Sum

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问题描述

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 :

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Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

Constraints:

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

解答

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/*
 * @Description: 15. 3Sum
 * @Author: libk
 * @Github: https://github.com/libk
 */
/**
 * @param {number[]} nums
 * @return {number[][]}
 */
const threeSum = function (nums) {
  const res = []

  if (!nums || nums.length < 3) {
    return res
  }

  nums.sort((a, b) => a - b)

  for (let i = 0; i < nums.length; i++) {
    if (nums[i] > 0) {
      break
    }
    if (i > 0 && nums[i - 1] === nums[i]) {
      continue
    }
    let left = i + 1
    let right = nums.length - 1
    while (left < right) {
      let sum = nums[i] + nums[left] + nums[right]
      if (sum === 0) {
        res.push([nums[i], nums[left], nums[right]])
        left++
        right--
        while (left < right && nums[left] === nums[left - 1]) {
          left++
        }
        while (left < right && nums[right] === nums[right + 1]) {
          right--
        }
      } else if (sum < 0) {
        left++
      } else {
        right--
      }
    }
  }
  return res
}