[Leetcode] Reverse Linked List II (Java)

最新推荐文章于 2025-09-11 19:19:55 发布

草鞋大脚

最新推荐文章于 2025-09-11 19:19:55 发布

阅读量571

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/StarCXDJ/article/details/18406933

Leetcode 专栏收录该内容

150 篇文章

订阅专栏

本文介绍了一种在链表中从位置m到n进行逆序的方法，并提供了一个Java实现示例。该方法采用原地修改的方式，仅需一次遍历即可完成逆序操作。

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

规定m，n的链表逆序

public class Solution {
  	public void reverse(ListNode pre,ListNode next){
		ListNode last = pre.next;
		ListNode cur = last.next;
		while(cur!= next){
			last.next = cur.next;
			cur.next = pre.next;
			pre.next = cur;
			cur = last.next;
		}
	}
	
	public ListNode reverseBetween(ListNode head, int m, int n) {
		ListNode res = new ListNode(-1);
		res.next = head;
		int index = 1;
		ListNode tmp = res;
		while(tmp!=null&&index<m){
			tmp = tmp.next;
			index++;
		}
		ListNode pre = tmp;
		while(tmp!=null&&index<n){
			tmp = tmp.next;
			index++;
		}
		tmp=tmp.next;
		ListNode next=tmp.next;
		reverse(pre, next);
		return res.next;
	}
}