LeetCode:Palindrome Linked List

Palindrome Linked List

Total Accepted: 28267 Total Submissions: 113712 Difficulty: Easy

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?

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思路：

1.用一个“快”指针(fast)和“慢”指针(slow)，同时后移。两个指针的后移速度是：fast = 2*slow，即slow走一步，fast走两步。

2.slow指针最终的位置链表的中间位置。

3.将前半部分链表逆置。

4.比较两个部分。

code:

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public boolean isPalindrome(ListNode head) {        if(head == null) {            return true;        }        ListNode slow = head;        ListNode fast = head;        ListNode p = slow.next;        ListNode pre = slow;        //find mid pointer, and reverse head half part        while(fast.next != null &&fast.next.next != null) {            fast = fast.next.next;            pre = slow;            slow = p;            p = p.next;            slow.next = pre;        }        //odd number of elements, need left move slow one step        if(fast.next == null) slow = slow.next;            slow = slow.next;            p = p.next;        }        return true;    }}