move-zeroes.md

283. Move Zeroes - 移动零

Tags - 题目标签

Description - 题目描述

EN:

Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.

Note that you must do this in-place without making a copy of the array.

 

Example 1:

Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]

Example 2:

Input: nums = [0]
Output: [0]

 

Constraints:

• 1 <= nums.length <= 104
• -231 <= nums[i] <= 231 - 1

 

Follow up: Could you minimize the total number of operations done?

ZH-CN:

给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。

请注意 ，必须在不复制数组的情况下原地对数组进行操作。

 

示例 1:

输入: nums = [0,1,0,3,12]
输出: [1,3,12,0,0]

示例 2:

输入: nums = [0]
输出: [0]

 

提示:

• 1 <= nums.length <= 104
• -231 <= nums[i] <= 231 - 1

 

进阶：你能尽量减少完成的操作次数吗？

Link - 题目链接

LeetCode - LeetCode-CN

Latest Accepted Submissions - 最近一次 AC 的提交

Language	Runtime	Memory	Submission Time
typescript	88 ms	46.5 MB	2023/03/25 21:27

/**
 Do not return anything, modify nums in-place instead.
 */
function moveZeroes(nums: number[]): void {
  let endWithZero = nums.findIndex(i => i === 0);

  if (endWithZero === -1) {
    return;
  }

  for (let i = endWithZero + 1; i < nums.length; i++) {
    if (nums[i] !== 0) {
      nums[endWithZero] = nums[i];
      endWithZero += 1;
    }
  }


  for (; endWithZero < nums.length; endWithZero++) {
    nums[endWithZero] = 0;
  }
};

My Notes - 我的笔记

两个指针一前一后

思路：一前一后两个指针，前指针先找到第一个0的位置，后指针从前指针+1开始遍历； 遇到0不处理，遇到非0则将后指针所在值赋给前指针所在值，并前指针+1; 后指针遍历完数组，将前指针后的数字全赋0即可。

/**
 Do not return anything, modify nums in-place instead.
 */
function moveZeroes(nums: number[]): void {
  let endWithZero = nums.findIndex(i => i === 0);

  if (endWithZero === -1) {
    return;
  }

  for (let i = endWithZero + 1; i < nums.length; i++) {
    if (nums[i] !== 0) {
      nums[endWithZero] = nums[i];
      endWithZero += 1;
    }
  }


  for (; endWithZero < nums.length; endWithZero++) {
    nums[endWithZero] = 0;
  }
};