问题描述
Given a singly linked list, determine if it is a palindrome.
Example 1:
Input: 1->2
Output: false
Example 2:
Input: 1->2->2->1
Output: true
Follow up:
Could you do it in O(n) time and O(1) space?
思路一:
把链表里的数存进一个数组里,然后在数组里利用索引判断。
思路二:
利用栈的后进先出原则,把链表后半段的数压栈,一边从栈里pop一边与链表前半段的数判断
但是,前两种方法的空间复杂度均不符合题目要求
思路三:
找到链表中点后,利用快指针fast,慢指针slow,把链表后半段反转,即指针由后指向前一个数,然后两头一起判断,代码如下:
/**
* 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 || head->next==NULL) return true;
ListNode *slow = head,*fast = head;
while(fast->next && fast->next->next)
{
slow = slow->next;
fast = fast->next->next;
}
ListNode *mid = slow->next,*pre=head;
while(mid->next)
{
ListNode *flag = mid->next;
mid->next= flag->next;
flag->next =slow->next;
slow->next=flag;
}
while(slow->next)
{
slow = slow->next;
if(pre->val != slow->val)
return false;
pre=pre->next;
}
return true;
}
};