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?

O(n) time, O(1) extra space

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/**
 * 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) {
    	// Note: the input will be changed, which is considered as a bad practice
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy;
        // The above 4 lines can be replaced by the following 3 lines
        // if (head == null) return true;
        // ListNode slow = head;
        // ListNode fast = head.next;
        
        // find the middle point
        while (fast != null && fast.next !=null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        // reverse the latter half, then compare the value from two ends to the middle point
        ListNode end = reverseLinkedList(slow.next);
        while (head != null && end != null) {
            if (head.val != end.val) {
                return false;
            } else {
                head = head.next;
                end = end.next;
            }
        }
        return true;
    }
    
    public ListNode reverseLinkedList(ListNode head) {
        ListNode prev = null;
        while (head != null) {
            ListNode next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
}