Leetcode刷题记录——206. 反转链表

我实现了两个迭代 和 递归
题目比较简单：
原链表为
1->2->3->4->5->NULL
操作后为
5->4->3->2->1->NULL
这道题要注意：一定别忘了将原输入头结点的next设为None

迭代：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if head is None:
            return [] 
        
        thisstack = []
        print(thisstack)
        while  True:#head is not None:
            thisstack.append(head)
            #print(head.val)
            if head.next is not None:
                head = head.next#thisstack[-1]
            else:
                break
        #print(thisstack)
        tempnode = head#tempnode
        while thisstack != []:
            tempnode.next = thisstack.pop(-1)
            tempnode = tempnode.next
        tempnode.next = None
        return head

递归：

class Solution:
    def __init__(self):
        self.newhead = None
    def reverseList(self, head: ListNode) -> ListNode:
        if head is None:
            return [] 
        self.func(head)
        head.next = None
        return self.newhead
    def func(self,node):
        if  node.next is not None:
            self.func(node.next)
            node.next.next = node
        else:
            self.newhead = node