LeetCode 234. Palindrome Linked List （回文链表）

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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题目标签：Linked List

　　题目给了我们一个 linked list，让我们判断它是不是回文。

　　这里可以利用 #206 Reverse Linked List 把右边一半的链表 倒转，然后从左右两头开始比较链表是否是回文。

　　这样的话，首先要找到链表的中间点，然后开始倒转右半边链表。

　　可以利用slow fast pointers 来找到中间点，当fast 走完的时候，slow 正好在中间。

 

 

Java Solution:

Runtime beats 39.01% 

完成日期：06/11/2017

关键词：singly-linked list

关键点：利用slow fast 指针来找到中间点

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution 
{
    public boolean isPalindrome(ListNode head) 
    {
        ListNode slow = head;
        ListNode fast = head;
        ListNode tail;
        
        // find the middle
        while(fast != null && fast.next != null)
        {
            slow = slow.next;
            fast = fast.next.next;
        }
        
        if(fast != null) // odd length
            slow = slow.next; // move middle to right half
        
        // reverse right half
        tail = reverse(slow);
        
        // compare head and tail
        while(tail != null)
        {
            if(head.val != tail.val)
                return false;
            
            head = head.next;
            tail = tail.next;
        }
        
        return true;
    }
    
    private ListNode reverse(ListNode cursor)
    {
        ListNode pre = null;
        ListNode next;
        
        while(cursor != null)
        {
            next = cursor.next;
            cursor.next = pre;
            pre = cursor;
            cursor = next;
        }
        
        return pre; // tail node
    }
}

参考资料：https://discuss.leetcode.com/topic/33376/java-easy-to-understand

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