Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

给一个链表，判断是否为回文。

如果是字符串就比较容易判断。但链表不能通过index来直接访问，而只能从头开始遍历到某个位置。

解法1: 用fast和slow两个指针，每次快指针走两步，慢指针走一步，等快指针走完时，慢指针的位置就是中点。还需要用栈，每次慢指针走一步，都把值存入栈中，等到达中点时，链表的前半段都存入栈中了，由于栈的后进先出的性质，就可以和后半段链表按照回文对应的顺序比较。

解法2: O(1) sapce，不能使用stack了，而是将链表中的一半翻转一下，这样前后两段链表就可以按照回文的顺序比较。

Python：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#

class Solution:
# @param {ListNode} head
# @return {boolean}
def isPalindrome(self, head):
reverse, fast = None, head
# Reverse the first half part of the list.
while fast and fast.next:
fast = fast.next.next
head.next, reverse, head = reverse, head, head.next

# If the number of the nodes is odd,
# set the head of the tail list to the next of the median node.
tail = head.next if fast else head

# Compare the reversed first half list with the second half list.
# And restore the reversed first half list.
is_palindrome = True
while reverse:
is_palindrome = is_palindrome and reverse.val == tail.val
reverse.next, head, reverse = head, reverse, reverse.next
tail = tail.next

return is_palindrome　　

C++: 解法1

class Solution {
public:
bool isPalindrome(ListNode* head) {
if (!head || !head->next) return true;
ListNode *slow = head, *fast = head;
stack<int> s;
s.push(head->val);
while (fast->next && fast->next->next) {
slow = slow->next;
fast = fast->next->next;
s.push(slow->val);
}
if (!fast->next) s.pop();
while (slow->next) {
slow = slow->next;
int tmp = s.top(); s.pop();
if (tmp != slow->val) return false;
}
return true;
}
};

C++: 解法2

class Solution {
public:
bool isPalindrome(ListNode* head) {
if (!head || !head->next) return true;
ListNode *slow = head, *fast = head;
while (fast->next && fast->next->next) {
slow = slow->next;
fast = fast->next->next;
}
ListNode *last = slow->next, *pre = head;
while (last->next) {
ListNode *tmp = last->next;
last->next = tmp->next;
tmp->next = slow->next;
slow->next = tmp;
}
while (slow->next) {
slow = slow->next;
if (pre->val != slow->val) return false;
pre = pre->next;
}
return true;
}
};

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