标签：ext ref lis list www http next 参考 递归
题目描述
输入一个链表，反转链表后，输出新链表的表头。
思路一
使用头插法迭代进行反转
实现
/* public class ListNode { int val; ListNode next = null; ListNode(int val) { this.val = val; } }*/ public class Solution { public ListNode ReverseList(ListNode head) { ListNode newHead = new ListNode(-1); while(head != null){ ListNode tmp = head.next; head.next = newHead.next; newHead.next = head; head = tmp; } return newHead.next; } }
思路二
递归
实现
public ListNode ReverseList(ListNode head) { if (head == null || head.next == null) return head; ListNode next = head.next; head.next = null; ListNode newHead = ReverseList(next); next.next = head; return newHead; }
思路参考：https://www.nowcoder.com/discuss/198840
标签：ext ref lis list www http next 参考 递归
原文地址：https://www.cnblogs.com/MWCloud/p/11323016.html
