234 回文鍊錶

請判斷乙個鍊錶是否為回文鍊錶。

示例 1:

輸入: 1->2

輸出: false

輸入: 1->2->2->1

輸出: true

你能否用 o(n) 時間複雜度和 o(1) 空間複雜度解決此題？

思路1: 借助棧依次壓入每個節點，然後判斷棧頂和當前鍊錶節點的每個值。

# definition for singly-linked list.

# class listnode:
# def __init__(self, val=0, next=none):
# self.val = val
# self.next = next
class
solution
:def
ispalindrome
(self, head: listnode)
->
bool
: stack =
cur = head
while cur is
notnone
: cur = cur.
next
while head is
notnone
:if head.val != stack.pop(
).val:
return
false
head = head.
next
return
true

# definition for singly-linked list.

# class listnode:
# def __init__(self, val=0, next=none):
# self.val = val
# self.next = next
class
solution
:def
ispalindrome
(self, head: listnode)
->
bool
: stack =
cur = head
right = head.
next
while cur.
next
isnot
none
and cur.
next
.next
isnot
none
: right = right.
next
cur = cur.
next
.next
while right is
notnone
: right = right.
next
while stack:
if head.val != stack.pop(
).val:
return
false
head = head.
next
return
true

# definition for singly-linked list.

# class listnode:
# def __init__(self, val=0, next=none):
# self.val = val
# self.next = next
class
solution
:def
ispalindrome
(self, head: listnode)
->
bool
:if head is
none
or head.
next
isnone
:return
true
n1 = head
n2 = head
while n2.
next
isnot
none
and n2.
next
.next
isnot
none
: n1 = n1.
next
n2 = n2.
next
.next
n2 = n1.
next
n1.next
=none
n3 =
none
# 反轉右半部分鍊錶
while n2 is
notnone
: n3 = n2.
next
n2.next
= n1
n1 = n2
n2 = n3
# 進行判斷是否為回文結構
n3 = n1 #儲存最後1個節點
n2 = head
res =
true
while n1 is
notnone
and n2 is
notnone
:if n1.val != n2.val:
res =
false
break
n1 = n1.
next
n2 = n2.
next
# 將鍊錶復原
n1 = n3.
next
n3.next
=none
while n1 is
notnone
: n2 = n1.
next
n1.next
= n3
n3 = n1
n1 = n2
return res

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