leetcode 206. 反转链表 python_手撕代码目leetcode 206、反转链表python

题目描述：

 题解一：

class Solution(object):
    def reverseList(self, head):
        if head==None or head.next==None:
            return head
        pre = None
        cur = head
        nxt = cur.next
        while nxt:
            cur.next = pre
            pre = cur
            cur = nxt
            nxt = nxt.next
        cur.next = pre
        return cur 

题解二：递归

1.递归终止条件：输入head为空或head.next为空，返回head

2.递归返回值：反转后的head

3.当前递归：

将head->head.next变为head.next->head

对head.next调用reverseList，得到head.next反转后的结果。

head.next.next变为head

head.next设为空

class Solution(object):
    def reverseList(self, head):
        if head==None or head.next==None:
            return head
        node  = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return node