234. Palindrome Linked List

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?
分析
首先要尋找中點，原理是使用快慢指針，每次快指針走兩步，慢指針走一步。同時還要用棧，每次慢指針走一步，都把值存入棧中。等快指針走完時，鏈表的前半段都存入棧中了。最後慢指針繼續往前走，每次與棧頂元素進行比較。空間複雜度O(n)。
如何做到用O(1)空間呢？可以先找到中點，把後半段reverse一下，然後比較兩個小鏈表。

// Palindrome Linked List
// Time Complexity: O(n), Space Complexity: O(1)
public class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head == null) return true;

        final ListNode middle = findMiddle(head);
        middle.next = reverse(middle.next);

        ListNode p1 = head;
        ListNode p2 = middle.next;
        while (p1 != null && p2 != null && p1.val == p2.val) {
            p1 = p1.next;
            p2 = p2.next;
        }
        return p2 == null;

    }
    private static ListNode findMiddle(ListNode head) {
        if (head == null) return null;

        ListNode slow = head;
        ListNode fast = head.next;

        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }

    private static ListNode reverse(ListNode head) {
        ListNode prev = null;

        while (head != null) {
            ListNode tmp = head.next;
            head.next = prev;
            prev = head;
            head = tmp;
        }

        return prev;
    }
}