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描述
Given the head of a singly linked list, return true if it is a palindrome.
Example 1:
Input: head = [1,2,2,1]
Output: true
Example 2:
Input: head = [1,2]
Output: false
Note:
The number of nodes in the list is in the range [1, 10^5].
0 <= Node.val <= 9
解析
根据题意，就是判断一个列表中包含的值，是否是回文序列。简单的方法就是遍历找出链表中的数放到列表 r 中，然后遍历判断 r[i] 和 r[n-i-1] 是否相等，如果不想等直接返回 False ，否则遍历结束直接返回 True 。
解答
class ListNode(object):
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution(object):
def isPalindrome(self, head):
"""
:type head: ListNode
:rtype: bool
"""
if not head.next:
return True
r = [head.val]
while head.next:
head = head.next
r.append(head.val)
n = len(r)
for i in range(n//2):
if r[i]!=r[n-i-1]:
return False
return True
运行结果
Runtime: 1120 ms, faster than 51.01% of Python online submissions for Palindrome Linked List.
Memory Usage: 86.2 MB, less than 10.38% of Python online submissions for Palindrome Linked List.
解析
其实上面的过程还可以简化一下，只需要将得到的链表中的值组成的列表，直接进行逆向比对即可判断是否为回文链表。
解答
class ListNode(object):
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution(object):
def isPalindrome(self, head):
"""
:type head: ListNode
:rtype: bool
"""
result = []
while head:
result.append(head.val)
head = head.next
return result == result[::-1]
运行结果
Runtime: 1048 ms, faster than 68.12% of Python online submissions for Palindrome Linked List.
Memory Usage: 85.5 MB, less than 17.93% of Python online submissions for Palindrome Linked List.
解析
上面两种解法在本质上没有区别，可以看出两者的运行速度都相当慢，因为有一部分时间都消耗在了组织新的列表上了，最省事的解法肯定是一次遍历。思路如下：
- 定位到链表的中间点
- 然后将后半部分的链表节点进行逆序排列成新的链表
- 将新的链表与前半部分进行比较，相等即为回文链表
很明显运行速度和所用内存都有了明显提升。
解答
class ListNode(object):
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution(object):
def isPalindrome(self, head):
"""
:type head: ListNode
:rtype: bool
"""
if not head or not head.next:
return True
def reverseList(self, head):
newhead=None
while head:
p=head
head=head.next
p.next=newhead
newhead=p
return newhead
slow=fast=head
while fast and fast.next:
fast=fast.next.next
slow=slow.next
if fast:
slow=slow.next
newhead=reverseList(self,slow)
while newhead:
if newhead.val!=head.val:
return False
newhead=newhead.next
head=head.next
return True
运行结果
Runtime: 1016 ms, faster than 74.84% of Python online submissions for Palindrome Linked List.
Memory Usage: 66.9 MB, less than 86.18% of Python online submissions for Palindrome Linked List.
原题链接：leetcode.com/problems/pa…
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