刷题记录-剑指offer24：反转链表

输入一个链表，反转链表后，输出新链表的表头。

法1：递归

public class Solution {
    public ListNode ReverseList(ListNode head) {
        if(head == null||head.next == null)
            return head;
        ListNode next = head.next;
        ListNode newHead = ReverseList(next);
        next.next = head;
        head.next = null;
        return newHead;
    }
}

递归的结束条件是遍历到原链表的尾节点，最重要的是next.next = head，实现链表反转

法2：头插法

public class Solution {
    public ListNode ReverseList(ListNode head) {
        if(head == null||head.next == null)
            return head;
        ListNode newList = new ListNode(-1);
        while(head!=null){
            ListNode next = head.next;
            head.next = newList.next;
            newList.next = head;
            head = next;
        }
        return newList.next;

    }
}