力扣labuladong——一刷day17

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前言

判断是否是回文链表，可以用栈，或者递归，时间复杂度和空间复杂度都是on，可以使用快慢指针找到链表中点，然后反转后一半链表，空间复杂度降为o1

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一、力扣234. 回文链表

用栈实现

/**
 * 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<ListNode> deq = new ArrayDeque<>();
        ListNode p = head;
        while(p != null){
            deq.offerLast(p);
            p = p.next;
        }
        p = head;
        while(!deq.isEmpty()){
            ListNode temp = deq.pollLast();
            if(temp.val != p.val){
                return false;
            }
            p = p.next;
        }
        return true;
    }
}

二、力扣234. 回文链表

递归后序实现

/**
 * 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 {
    ListNode left;
    public boolean isPalindrome(ListNode head) {
        left = head;
        return fun(head);
    }
    public boolean fun(ListNode head){
        if(head == null){
            return true;
        }
        boolean flag = fun(head.next);
        flag = flag && head.val == left.val;
        left = left.next;
        return flag;
    }
}