206. 反转链表

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。

示例 1：

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]

示例 2：

输入：head = [1,2]
输出：[2,1]

示例 3：

输入：head = []
输出：[]

提示：

• 链表中节点的数目范围是 [0, 5000]
• -5000 <= Node.val <= 5000

进阶：链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？

/**
 * 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 {
    public ListNode reverseList(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode first = head;
        ListNode cur = head.next;
        ListNode last = head.next.next;
        while(first != null){
            cur.next=last;
            if(last == head){
                last.next=null;
            }
            last=cur;
            cur=first;
            first=first.next;
        }
        cur.next=last;
        if(last == head){
            last.next=null;
        }
        return cur;
    }
}

 注意处理链表的head节点，如果不在最后再验证一下if last == head这个情况，就会出现环路, 合并下这两种情况

class Solution {
    public ListNode reverseList(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode first = head;
        ListNode cur = head.next;
        ListNode last = head.next.next;
        while(first != null){
            cur.next=last;
            last=cur;
            cur=first;
            first=first.next;
        }
        cur.next=last;
        head.next=null;
        return cur;
    }
}