输入一个链表的头节点,从尾到头反过来返回每个节点的值(用数组返回)。
示例 1:
输入:head = [1,3,2]
输出:[2,3,1]
限制:
0 <= 链表长度 <= 10000
作者:Krahets
链接:https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5dt66m/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reversePrint(self, head: ListNode) -> List[int]:
ls=[]
while(head):
ls.insert(0,head.val)
head=head.next
return ls
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reversePrint(self, head: ListNode) -> List[int]:
ls=[]
while(head):
ls.append(head.val)
head=head.next
ls.reverse()
return ls
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
# 递归
class Solution:
def reversePrint(self, head: ListNode) -> List[int]:
if not head:
return []
else:
return self.reversePrint(head.next)+[head.val]