题目

原始地址：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode reverseBetween(ListNode head, int m, int n) {            }}

描述

局部翻转一个单链表，从m到n。要求原地调整并且只能遍历一次。

分析

这个题目的思路比较好理解，首先找到第m-1个节点，然后反转第m到n个节点，最后再将前m-1个节点，反转后的第m到第n节点以及n之后的节点分别连起来即可。

解法

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode reverseBetween(ListNode head, int m, int n) {        ListNode dummy = new ListNode(0), prev, curr, next = null, start = dummy, end;        dummy.next = head;        for (int i = 0; i < m - 1; i++) {            start = start.next;        }        end = start.next;        prev = null;        curr = start.next;        for (int i = m; i <= n; i++) {            next = curr.next;            curr.next = prev;            prev = curr;            curr = next;        }        start.next = prev;        end.next = next;        return dummy.next;    }}