题目
给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。
代码
/**
* 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; }
* }im
*/
class Solution {
public ListNode reverseList(ListNode head) {
ListNode cur = head;
ListNode pre = null;
while(cur != null) {
ListNode next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
return pre;
}
}
结论
- 虚拟出一个pre节点很关键
- 注意停止条件
- 注意交换引用的逻辑
