Problem: 92. 反转链表 II

👨‍🏫 参考题解

• 特殊情况
  

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    ListNode reverseBetween(ListNode head, int left, int right) {
        ListNode dummy = new ListNode(0, head);// 
        ListNode p0 = dummy;
        for(int i = 0; i < left - 1; i++)
            p0 = p0.next;
        ListNode pre = null;
        ListNode cur = p0.next; // 局部链表的第一个节点
        // 循环反转局部的链表
        for(int i = 0; i < right - left + 1; i++)
        {
            ListNode nxt = cur.next;
            cur.next = pre;
            pre = cur;
            cur = nxt;
        }
        // 处理局部链表的首尾与整体链表的衔接
        p0.next.next = cur;// p0.next 是局部链表的第一个节点
        p0.next = pre;//pre 局部链表的最后一个节点
        return dummy.next;
    }
}