20210406.md

Algorithm

234. Palindrome Linked List

Description

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

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?

Solution

/**
 * 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) {
        ListNode fast = head, slow = head;
        while(fast!=null && fast.next!=null){
            fast = fast.next.next;
            slow = slow.next;
        }
        // odd nodes: let right half smaller
        if (fast != null) {
            slow = slow.next;
        }
        slow = reverse(slow);
        fast = head;

        while(slow!=null){
            if(fast.val!=slow.val){
                return false;
            }
            fast = fast.next;
            slow = slow.next;
        }
        return true;
    }
    private ListNode reverse(ListNode head){
        ListNode prev = null;
        ListNode now = head;
        while(now!=null){
            ListNode next = now.next;
            now.next = prev;
            prev = now;
            now = next;
        }
        return prev;
    }
}

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方法名称	位置	类型	作用	线程数量	调用次数
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start	Thread类中	同步方法	负责启动线程	会产生新线程	一次