LeetCode Remove Element

最新推荐文章于 2025-08-23 14:22:20 发布

原创最新推荐文章于 2025-08-23 14:22:20 发布 · 213 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

leetcode 专栏收录该内容

124 篇文章

订阅专栏

本文介绍了一种在数组中高效移除特定值的方法，并保持内存使用不变。通过双指针技术实现，确保了操作的原地完成。文章提供了一个具体的示例，展示了如何将指定值从数组中移除，并返回更新后的数组长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

没啥子思路可言，通过设置首位两个索引变量即可.

代码如下：

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int len = nums.size();
        int i=0,j=len-1;
        while(i<=j)
        {
            while(nums[i] != val && i < len)
            {
                i++;
            }
            while(nums[j] == val && j >= 0 )
            {
                j--;
            }
            if(i<j)
            {
                nums[i] = nums[i]^nums[j];
                nums[j] = nums[i]^nums[j];
                nums[i] = nums[i]^nums[j];
            }
        }
        return i;
    }
};