leetcode 206 反轉鍊錶

反轉乙個單鏈表。

示例:輸入: 1->2->3->4->5->null
輸出: 5->4->3->2->1->null
高階:你可以迭代或遞迴地反轉鍊錶。你能否用兩種方法解決這道題？

#棧

# 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:
ls=while(head):#壓棧
head=head.next
p1=p2=listnode(0,none)
while(ls):
t=ls.pop()
p2.next=t
p2=p2.next
p2.next=none    
return p1.next

# 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:
ls=if not head or not head.next :
return head
while(head):
head=head.next
p=p1=ls.pop()
while(ls):
p.next=ls.pop()
p=p.next
p.next=none
return p1

# 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:
p=none#建立空節點
while(head):
t=head.next#當前節點的下乙個
head.next=p#當前節點的next
p=head#後移
head=t#後移
return p

# 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:
#遞迴if not head or not head.next:
return head
t=self.reverselist(head.next)
head.next.next=head
head.next=none
return t

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