标签:inpu node ref == return false 注意 isp com
Description: Given a singly linked list, determine if it is a palindrome.
Link: https://leetcode.com/problems/palindrome-linked-list/
Examples:
Example 1: Input: 1->2 Output: false Example 2: Input: 1->2->2->1 Output: true Example 3: Input: 0->0 Output: true
思路: 回文检测,注意保存node.val
class Solution(object): def isPalindrome(self, head): """ :type head: ListNode :rtype: bool """ if not head: return True p = head nodes = [] while p: nodes.append(p.val) p = p.next n = len(nodes) if n % 2 == 0: half = int(n/2) else: half = int(n/2)+1 i = 0 while i < half: if nodes[i] != nodes[n-i-1]: break i += 1 if i < half: return False return True
日期: 2020-12-01
Leetcode 234. Palindrome Linked List
标签:inpu node ref == return false 注意 isp com
原文地址:https://www.cnblogs.com/wangyuxia/p/14068799.html