LeetCode-Reverse Linked List

Problem:
Reverse a singly linked list.
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Hint:
A linked list can be reversed either iteratively or recursively. Could you implement both?
Analysis:
单链表逆置，两种方法，递归和迭代。

• 递归是如果head不是空或者head.next非空，将head.next递归逆置，然后将head和head.next换位置，并处理好head.next=null.
• 迭代基本会有一个pre和一个p以及一个temp节点q。首先确定pre,赋值p,赋值pre.next,申明节点q。然后循环：赋值q,赋值p.next=pre,pre=p(移动pre到最当前节点),p=q(移动p到下一个节点)。返回pre.

Anwser1(递归):

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode reverseList(ListNode head) {        if(head == null) return null;        if(head.next==null) return head;        else {            ListNode temp = head.next;            ListNode n = reverseList(temp);            temp.next = head;            head.next=null;            return n;        }    }}

Anwser2(迭代):

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode reverseList(ListNode head) {        if(head == null) return null;        if(head.next==null) return head;        ListNode pre=head;        ListNode p= pre.next;        pre.next = null;        ListNode q ;        while(p!=null){            q = p.next;            p.next = pre;            pre = p;            p = q;        }        return pre;    }}