leetcode hot 19

合集 - leetcode hot100(25)

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19.leetcode hot 1903-04

20.leetcode hot 2003-0521.leetcode hot 2103-0522.leetcode hot 2203-1223.leetcode hot 2303-1224.leetcode hot 2403-1225.leetcode hot 2503-12

收起

解题思路:这题思路主要是用某个容器（数组等）来存储链表，然后再用双指针或者同时遍历的方法就可以判断。我使用栈进行存储，出栈的元素相当于链表倒序，如果和链表正序的结果相同就证明是回文。如果想用O(1)的空间，就把前一半或后一半的链表倒转，然后遍历就可以了。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        Deque<Integer> deque = new LinkedList<>();
        ListNode p = head;
        while(p!=null){
            deque.push(p.val);
            p=p.next;
        }
        ListNode q = head;
        while(q!=null){
            if(q.val!=deque.peek()) return false;
            deque.pop();
            q = q.next;
        }
        return true;


    }
}