题解 | #反转链表#

相当于使用栈进行输出

# class ListNode:
#     def __init__(self, x):
#          self.val = x
#          self.next = None

# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @return ListNode类
#
class Solution:
    def ReverseList(self , head: ListNode) -> ListNode:
        s=[]
        m=None
        if head==None:
            return m
        while head:
            s.append(head)
            head=head.next
        s=s[::-1]
        num = len(s)
        for i in range(num):
            if i==num-1:
                s[i].next=None
                break
            s[i].next=s[i+1]
        return s[0]