力扣面试150 反转链表 II 三指针

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;}
}