[LeetCode]--234. Palindrome Linked List

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?

我的思路很简单，先遍历一遍找到长度，然后用堆保存着前面一半的数据然后与后面的对比，不一样就false。

/**
 * 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) {
        if (head == null || head.next == null)
            return true;
        ListNode p1 = head;
        Stack<Integer> stack = new Stack<Integer>();
        int len = 0, i = 0;
        while (p1 != null) {
            len++;
            p1 = p1.next;
        }
        p1 = head;
        while (i++ < len / 2) {
            stack.push(p1.val);
            p1 = p1.next;
        }
        if (len % 2 != 0)
            p1 = p1.next;
        while (p1 != null) {
            if (p1.val != stack.pop())
                return false;
            p1 = p1.next;
        }
        return true;
    }
}

再提供一种方法，就是用反转来做。

public boolean isPalindrome(ListNode head) {
        if (head == null) {
            return true;
        }

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

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

        return p2 == null;
    }

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

        return slow;
    }

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

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

        return prev;
    }