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;
  }
};