leetcode-234-easy

Palindrome Linked List

Given the head of a singly linked list, return true if it is a palindrome or false otherwise.

Example 1:

Input: head = [1,2,2,1]
Output: true
Example 2:

Input: head = [1,2]
Output: false
Constraints:

The number of nodes in the list is in the range [1, 105].
0 <= Node.val <= 9
Follow up: Could you do it in O(n) time and O(1) space?

思路一：刚开始想用栈来实现，后面想想直接用双端队列，先把链表转换成双端队列，然后取队列两头进行对比

    public static boolean isPalindrome(ListNode head) {
        Deque<Integer> deque = new ArrayDeque<>();

        while (head != null) {
            deque.push(head.val);

            head = head.next;
        }

        while (deque.size() > 1) {
            if (!deque.pollFirst().equals(deque.pollLast())) {
                return false;
            }
        }

        return true;
    }

思路二：看了一下题解，有更好的算法，用快慢指针，反转链表的后半部分，最后直接从中间开始比较链表的值