题解 | #链表内指定区间反转# 一次遍历，简单明了

遍历一遍链表，记录下index为m-1和m的节点node1和node2，当m<index<=n时，当前节点指向前一个节点进行反转，最后index等于n时再利用记录下来的node1和node2完善结果。

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param head ListNode类
     * @param m int整型
     * @param n int整型
     * @return ListNode类
     */
    public ListNode reverseBetween (ListNode head, int m, int n) {
        // write code here
        int index = 1;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy;
        ListNode node1 = dummy;
        ListNode node2 = dummy;
        while (head != null) {
            ListNode next = head.next;
            if (index == m - 1) {
                node1 = head;
            }
            if (index == m) {
                node2 = head;
            }
            if (m < index && index <= n) {
                head.next = prev;
            }
            if (index == n) {
                node1.next = head;
                node2.next = next;
            }
            prev = head;
            head = next;
            index++;
        }
        return dummy.next;
    }
}