题目链接
    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：

    1. 输入：head = [1,2,3,4,5]
    2. 输出：[5,4,3,2,1]

    示例2：

    1. 输入：head = [1,2]
    2. 输出：[2,1]

    示例3：

    1. 输入：head = []
    2. 输出：[]

    提示：

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

    1. 双指针

      1. /**
      2. * Definition for singly-linked list.
      3. * public class ListNode {
      4. *     int val;
      5. *     ListNode next;
      6. *     ListNode() {}
      7. *     ListNode(int val) { this.val = val; }
      8. *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
      9. * }
      10. */
      11. class Solution {
      12.  public ListNode reverseList(ListNode head) {
      13.      ListNode pre = null;
      14.      ListNode current = head;
      15.      ListNode temp = null;
      16.      while (current != null){
      17.          temp = current.next;
      18.          current.next = pre;
      19.          pre = current;
      20.          current = temp;
      21.      }
      22.      return pre;
      23.  }
      24. }

    2. 递归

      /**
* 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) {
     return reverse(null, head);
 }
 public ListNode reverse(ListNode prev, ListNode cur) {
     // 递归终止条件，在当前指向null时
     // 空链表 head=null 或者是 已经翻转完最后一个元素
     if(cur == null) {
         return prev;
     }
     ListNode temp = cur.next; //保存下一个节点
     cur.next = prev; // 翻转当前节点
     return reverse(cur, temp); // 相当于双指针的 pre = current; current = temp;
 }
}