题目：
反转一个单链表。
示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-linked-list
（1）递归
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
ListNode res = null;
public ListNode reverseList(ListNode head) {
if(head==null) {
return null;
}
ListNode q = head.next;
ListNode p = head;
p.next = null;
fReverseList(q,p);
return res;
}
private void fReverseList(ListNode head, ListNode pre) {
if(head==null) {
res = pre;
return;
}
ListNode p = head.next;
head.next = pre;
fReverseList(p,head);
}
}
（2）非递归
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
if(head==null) {
return null;
}
ListNode pi = head;
ListNode pj = pi.next;
if(pj==null) {
return pi;
}
ListNode pk = pj.next;
if(pk==null) {
pi.next = null;
pj.next = pi;
return pj;
}
pi.next = null;
while(pk!=null) {
pj.next = pi;
pi = pj;
pj = pk;
pk = pk.next;
}
pj.next = pi;
return pj;
}
}
