先找到链表中点,然后将右半部分链表反转,再判断左右两个链表节点值是否相等。 classSolution{public:boolisPalindrome(ListNode* head){if(!head || !head->next)returntrue;// find middle nodeListNode *slow = head, *fast = head->next;while(fast && fast->next) { slow = slow->next; fast = fast...
这一点太弱了。 booljudge(structListNode *p,structListNode *q) {while( p &&q) {if(p->val != q->val)returnfalse; p= p->next; q= q->next; }returnp==NULL && q==NULL; }boolisPalindrome_using_count(structListNode*head) {inti,n =0;structListNode * p = head, *last = NULL, *...
⭐ Leetcode 解題紀錄 ⭐題型資料結構Python SolutionC++ SolutionNote ⭐BFS 相關題型 ⭐ 104 Maximum Depth of Binary Tree BFS (分層) Python 94 Binary Tree Inorder Traversal BFS (分層) Tree Python 內含 處理 Tree 樹問題的重點 102 Binary Tree Level Order Traversal BFS (分層) Tree Python ...
Can you solve this real interview question? Palindrome Linked List - Given the head of a singly linked list, return true if it is a palindrome or false otherwise. Example 1: [https://assets.leetcode.com/uploads/2021/03/03/pal1linked-list.jpg] Inpu
LeetCode: 234. Palindrome Linked List 题目描述 Given a singly linked list, determine if it is a palindrome. Example 1: Input: 1->2 Output: false 1. 2. Example 2: Input: 1->2->2->1 Output: true 1. 2. Follow up: Could you do it in O(n) time and ...
Leetcode 234 Palindrome Linked List 复杂度为时间O(n) 和空间(1)解法,1.问题描写叙述给定一个单链表,推断其内容是不是回文类型。比如1–>2–>3–>2–>1。时间和空间复杂都尽量低。2.方法与思路1)比較朴素的算法。因为给定的数据结构是单链表,要訪问链表的尾部元素,
Given a singly linked list, determine if it is a palindrome.Example 1:Input: 1->2Output: falseExample 2:Input: 1->2->2->1Output: trueFollow up:Could you do it in O(n) time and O(1) space?来源:力扣(LeetCode)链接:https://leetcode-cn.com/problems/palindrome-linked-list著作权归领...
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? 反转链表 复杂度 时间O(N) 空间 O(1) 思路 两个指针都从头出发,快指针每次两步,慢指针每次一步,这样快指针的下一个或下下个为空时,慢指针就在链...
LeetCode problem solving notes. Determine whether a linked list is a palindrome linked list. First convert the linked list into an array, and then judge whether the array is a palindrome array.
# Definitionforsingly-linked list.#classListNode:# def__init__(self,x):# self.val=x # self.next=NoneclassSolution(object):defisPalindrome(self,head):""":type head:ListNode:rtype:bool""" node_list=[]whilehead:node_list.append(head)head=head.nextforiinrange(int(len(node_list)/2)):...