234. Palindrome Linked List（重要）

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?

难点是单链表。

做法是用快慢指针找到链表的中心。然后将链表的后一半逆转。然后判断前后两个部分是否相等。

有个注意的地方，即下面的代码。当链表元素为奇数时，后一半比前一半少一个，所以while在pre为空时就停止，中心不用比较！

while (head&&pre){
if (head->val != pre->val){
return false;
}
else{
head = head->next;
pre = pre->next;
}
}

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool isPalindrome(ListNode* head) {
if (head == NULL){
return true;
}
ListNode* faster = head;
ListNode* slower = head;

while (faster->next&&faster->next->next){
faster = faster->next->next;
slower = slower->next;
}

ListNode* pre = NULL;
ListNode* p = slower->next;
while (p){
ListNode* next = p->next;
p->next = pre;
pre = p;
p = next;
}

while (head&&pre){
if (head->val != pre->val){
return false;
}
else{
head = head->next;
pre = pre->next;
}
}
return true;
}
};