27. Remove Element

最新推荐文章于 2024-02-22 13:25:37 发布

原创最新推荐文章于 2024-02-22 13:25:37 发布 · 439 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#java #LeetCode

LeetCode 专栏收录该内容

32 篇文章

订阅专栏

本文介绍了一种删除数组中特定元素并返回新长度的方法，使用双指针技巧优化空间复杂度至O(1)，并详细解释了实现过程及优化思路。

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.

Hint:

Try two pointers.
Did you use the property of "the order of elements can be changed"?
What happens when the elements to remove are rare?

public class Solution {
	public int removeElement(int[] nums, int val) {
		int index = 0, len = nums.length-1;
		while (index <= len) {
			if (nums[index] == val)
				nums[index] = nums[len--];
			else
				index++;
		}
		return index;
	}
}
//这个是符合条件的数组不动，一旦出现不符合条件的就把最后面的数值放进来，这样减少了数组的操作。

public class Solution {
    public int removeElement(int[] nums, int val) {
        int idx = 0;
        for (int i = 0; i < nums.length; i++){
            if (nums[i] != val){
                nums[idx++] = nums[i]; 
            }
        }
        return idx;
    }
}
//遍历一遍数组将符合条件的数组按原来的次序排列

我倾向于第一种算法~