Palindrome Linked List(lintcode 223)

Description

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

Example

Given 1->2->1, return true.

Interface

/**
 * 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
    }
}

Idea

The code destroy the original structure of the list. If you don't want to do that, reverse the second part back.

Solution

/**
 * 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) {
            return true;
        }
        ListNode mid = findMid(head);
        ListNode newHead = reverse(mid);
        while (newHead != null) {
            if (newHead.val != head.val) {
                return false;
            }
            newHead = newHead.next;
            head = head.next;
        }
        return true;
    }

    private ListNode findMid(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }

    private ListNode reverse(ListNode head) {
        ListNode newHead = null;
        while (head != null) {
            ListNode nxt = head.next;
            head.next = newHead;
            newHead = head;
            head = nxt;
        }
        return newHead;
    }
}