27. Remove Element

最新推荐文章于 2025-08-22 19:21:44 发布

原创最新推荐文章于 2025-08-22 19:21:44 发布 · 241 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #Remove Element

leetcode 专栏收录该内容

18 篇文章

订阅专栏

本文介绍两种在数组中移除指定值的有效方法，并通过示例进行说明。方法一采用单次遍历，将非目标元素向前移动；方法二利用双指针技巧，通过交换元素并调整指针来实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

一,题目

Given an array and a value, remove all instances of that value in place and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:
Given input array nums = [3,2,2,3], val = 3

Your function should return length = 2, with the first two elements of nums being 2.

二,思路

方法一:
从头到尾遍历一遍, 对于后面的元素如果不是被删除的元素则前移

方法二:
使用two pointer, left right left索引值不为被删除值,则++
否则将right索引处的值赋给left索引处的值,right--

三,代码

public int removeElement(int[] nums, int val) {
		if (nums == null || nums.length == 0) {
			return 0;
		}

		int pointer = 0;
		for (int i = 0, len = nums.length; i < len; i++) {
			if (nums[i] != val) {
				nums[pointer++] = nums[i];
			}
		}

		return pointer;
	}

	public  int removeElement2(int[] nums, int val) {
		if (nums == null || nums.length == 0) {
			return 0;
		}
		int pointer = nums.length - 1;
		int i = 0;
		while (i <= pointer) {
			if(nums[i] == val){
				nums[i] = nums[pointer];
				pointer--;
			}else{
				i++;
			}
		}
		return pointer+1;
	}