[lintcode medium]Palindrome Linked List

检查链表是否为回文

最新推荐文章于 2020-08-30 10:19:39 发布

转载最新推荐文章于 2020-08-30 10:19:39 发布 · 76 阅读

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原文链接：http://www.cnblogs.com/kittyamin/p/5031674.html

本文介绍了一种算法，用于判断给定的单链表是否为回文结构，在O(n)时间内完成，并且仅使用O(1)的空间复杂度。

Palindrome Linked List

Implement a function to check if a linked list is a palindrome.

Example

Given 1->2->1, return true

Challenge

Could you do it in O(n) time and O(1) space?

////

1\find out the medium index of Linked list

2\ reverse the right part of linked list in place

3\compare their value;

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**
     * @param head a ListNode
     * @return a boolean
     */
    public boolean isPalindrome(ListNode head) {
        // Write your code here
        if(head==null || head.next==null) return true;
        if(head.next.next==null)
        {
            if(head.val==head.next.val)
            return true;
            
            else return false;
        }
        
        ListNode fast=head;
        ListNode slow=head;
        
        while(fast.next!=null && fast.next.next!=null )
        {
            fast=fast.next.next;
            slow=slow.next;
        }
        
        ListNode head2=slow.next;
        ListNode pre=null;
        ListNode cur=head2;
        
        slow.next=null;
        while(cur!=null)
        {
            ListNode next=cur.next;
            cur.next=pre;
            pre=cur;
            cur=next;
        }
       
        head2=pre;
        
        ListNode p=head;
        ListNode q=head2;
        while(q!=null)
        {
            if(p.val==q.val)
            {
                p=p.next;
                q=q.next;
            }
            else
             return false;
        }
    
    return true;
        
    }

转载于:https://www.cnblogs.com/kittyamin/p/5031674.html