leetcode note--leetcode 27 Remove Element

最新推荐文章于 2025-09-09 20:07:09 发布

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本文介绍了一种在不使用额外空间的情况下从数组中移除特定值的算法，并提供了两种实现方法，一种通过双指针技巧交换元素，另一种采用计数方式来达到目的。

27. Remove Element

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Total Accepted: 159692
Total Submissions: 436588
Difficulty: Easy
Contributors: Admin

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.

方法一：自己写的

public class Solution {
    public int removeElement(int[] nums, int val) {
        int len = nums.length;
        int slow = 0;
        int fast = len-1;
        if(len==0)return 0;
        while(slow<len && fast>=0){
            while(slow<len && nums[slow]!=val)slow++;
            while(fast>=0 && nums[fast]==val)fast--;
            if(slow<fast){
                nums[slow] = nums[fast];
                nums[fast] = val;
            }else{
                return slow;
            }
        }
        return slow;
    }
}

time:9ms;

方法二：参考了别人的。利用了jianxi的方法。。。。

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

time:9ms