Leetcode之Palindrome Linked List 问题 问题描述: Given a singly linked list, determine if it is a palindrome. Follow up: Could you do it in O(n) time and O(1) space? 示例: 1->2->9->4->9->2->1 return true; 1->2-&... LeetC
Given a singly linked list of characters, write a function that returnstrueifthe given listispalindrome,elsefalse. 题解1 - 使用辅助栈 根据栈的特性(FILO),可以首先遍历链表并入栈(最后访问栈时则反过来了),随后再次遍历链表并比较当前节点和栈顶元素,若比较结果完全相同则为回文。 又根据回文的特性,实际上...
Could you do it in O(n) time and O(1) space? 解题报告: 1. 找到LinkedList 的中点 2. 翻转链表(从中点的下一个节点到链表终点)得到另一个链表 3. 依次比较两个节点的值是否相等。 1/**2* Definition for singly-linked list.3* public class ListNode {4* int val;5* ListNode next;6* ListNo...
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Palindrome Linked List (E)题目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?
Palindrome Linked List 链表代码人生 Given a singly linked list, determine if it is a palindrome. Follow up: Could you do it in O(n) time and O(1) space? 思路: 把链表一分为二。把右边的一半翻转,再逐个比对左右的链表就可以。 /**
234. Palindrome Linked List(重要) 链表快慢指针文章分类运维 Given a singly linked list, determine if it is a palindrome. Follow up: Could you do it in O(n) time and O(1) space? 难点是单链表。 做法是用快慢指针找到链表的中心。然后将链表的后一半逆转。然后判断前后两个部分是否相等。
next = None class Solution(object): def isPalindrome(self, head): """ :type head: ListNode :rtype: bool """ node_list=[] while head: node_list.append(head) head=head.next for i in range(int(len(node_list)/2)): print(node_list[i].val,node_list[-i-1].val) if node_list[...
Palindrome Linked List Description Given a singly linked list, determine if it is a palindrome. Example 1: Input: Output: Example 2: Input: Output: Follow up: Could you do it in O(n) time and O(1) space? 分析 题目的意思是:判断一个链表是否是...LeetCode - 234. Palindrome Linked ...
LeetCode - 234 - Palindrome Linked List Given a singly linked list, determine if it is a palindrome. Follow up: Could you do it in O(n) time and O(1) space? 判断一个链表是不是回文,我的方法是把前一半链表翻转,然后再同时扫描,判断值是否相等。时间复杂度O(n),至于空间复杂度,我觉得...