leetcode -- 92. Reverse Linked List II

最新推荐文章于 2025-09-13 14:08:34 发布

原创最新推荐文章于 2025-09-13 14:08:34 发布 · 104 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

算法专栏收录该内容

212 篇文章

订阅专栏

本文介绍了一种在单次遍历中反转链表指定区间的算法，适用于中等难度的链表操作问题。通过实例演示了如何从位置m到n反转链表，并提供了完整的AC代码实现。

题目描述

题目难度:Medium
Reverse a linked list from position m to n.
Do it in one-pass.
Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

AC代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if(m == n) return head;
        ListNode newHead = null;
        ListNode curNode = null;
        int i = 0;
        if(m == 1) { //从头节点开始
            newHead = new ListNode(0);
            curNode = head;
            ListNode end = head; //m - n之间的元素翻转完成之后的尾节点
            while(i < n - m + 1){
                curNode = head;
                head = head.next;
                curNode.next = newHead.next;
                newHead.next = curNode; 
                i++;
            }
            end.next = head;
            return newHead.next;
        }
        else{
            i = 0;
            newHead = head;
            while(i < m - 2) {
                head = head.next;
                i++;
            }
            ListNode start = head; // m 的前一个节点
            ListNode end = head.next; //m - n之间的元素翻转完成之后的尾节点
            head = head.next;
            ListNode fakeNode = new ListNode(0);
            curNode = head;
            i = 0;
            while(i < n - m + 1){
                curNode = head;
                head = head.next;
                curNode.next = fakeNode.next;
                fakeNode.next = curNode; 
                i++;
            }
            end.next = head;
            start.next = fakeNode.next;
            return newHead;
        }        }
}