标签:fas eve lin public rev pre head input 判断
Given a singly linked list, determine if it is a palindrome.
Example 1:
Input: 1->2 Output: false
Example 2:
Input: 1->2->2->1 Output: true
Follow up:
Could you do it in O(n) time and O(1) space?
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public boolean isPalindrome(ListNode head) { if(head == null || head.next == null) return true; //根据快慢指针求中点 ListNode slow = head; ListNode fast = head; while(fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } if(fast != null) slow = slow.next; //slow是中点,将中点反转后 slow = reverse(slow); fast = head; while(slow != null) { if(slow.val != fast.val) return false; slow = slow.next; fast = fast.next; } return true; } public ListNode reverse(ListNode head) { ListNode pre = null; ListNode next = null; while(head != null) { next = head.next; head.next = pre; pre = head; head = next; } return pre; } }
234. Palindrome Linked List【Easy】【判断链表是否回文】
标签:fas eve lin public rev pre head input 判断
原文地址:https://www.cnblogs.com/Roni-i/p/10442933.html