package LeetCode_LinkedList;

/**
 * 題目：
 *      Given a singly linked list, determine if it is a palindrome.
 *      Example 1:
 *          Input: 1->2
 *          Output: false
 *      Example 2:
 *          Input: 1->2->2->1
 *          Output: true
 *      Follow up:
 *          Could you do it in O(n) time and O(1) space?
 * 解題思路：
 *      找到連結串列中間節點，將連結串列分成兩段，然後將後段連結串列進行反轉，
 *      最後將反轉後的連結串列依次與前半段連結串列進行比較，確定是否相等。
 */
public class IsPalindrome_234_1016 {
    public boolean IsPalindrome(ListNode head) {
        if (head == null || head.next == null) {
            return true;
        }

        ListNode fast = head;
        ListNode slow = head;

        //將連結串列分成前後兩部分
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }

        if (fast != null) {
            slow = slow.next;
        }

        //反轉連結串列
        slow = Reverse(slow);
        fast = head;

        //依次比較前後半段連結串列節點的值是否相等
        while (slow != null) {
            if (slow.val != fast.val) {
                return false;
            }
            slow = slow.next;
            fast = fast.next;
        }
        return true;
    }

    public ListNode Reverse(ListNode head) {
        ListNode pre = null;
        ListNode node = head;
        while (node != null) {
            ListNode next = node.next;
            node.next = pre;

            pre = node;
            node = next;
        }
        return pre;
    }

    public static void main(String[] args) {
        ListNode node1 = new ListNode(1);
        ListNode node2 = new ListNode(2);
        //ListNode node3 = new ListNode(2);
        //ListNode node4 = new ListNode(1);

        node1.next = node2;
        //node2.next = node3;
        //node3.next = node4;

        IsPalindrome_234_1016 test = new IsPalindrome_234_1016();
        boolean result = test.IsPalindrome(node1);
        System.out.println(result);
    }
}