27. Remove Element

最新推荐文章于 2024-01-20 09:00:07 发布

快乐遇见忧伤

最新推荐文章于 2024-01-20 09:00:07 发布

阅读量271

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/GPFYCF521/article/details/80792133

LeetCode 专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一种在原地移除数组中特定值的方法，该方法不使用额外空间，并保持O(1)的额外内存复杂度。通过双指针技巧，将非目标值前移，实现快速移除。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目
想法
Code

题目

Given an array nums and a value val, 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 by modifying the input array in-place with O(1) extra memory.
The order of elements can be changed. It doesn’t matter what you leave beyond the new length.

example1:
Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.
It doesn't matter what you leave beyond the returned length.
example2:
Given nums = [0,1,2,2,3,0,4,2], val = 2,
Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.
Note that the order of those five elements can be arbitrary.
It doesn't matter what values are set beyond the returned length.

想法

对于顺序没有要求，就是将非val的值移动到数组的前面，但是对于等于val的值并没有要求，不需要保存
两种想法：将等于val的值移动到后面，将不等于val的值移动到前面，显然将不等于的移动到前面更方便
因为移到后面需要检查被交换的元素，因为还没被访问过，但是移动到前面就不用担心，因为已经被访问过，随意覆盖就可以

Code

public int removeElement(int[] nums, int val) {
    int i = 0;
    for (int j = 0; j < nums.length; j++) {
        if (nums[j] != val) {
            nums[i] = nums[j];
            i++;
        }
    }
    return i;
}