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...
奇数时,先pop掉中间一个。 1classSolution(object):2defisPalindrome(self, head):3"""4:type head: ListNode5:rtype: bool6"""7ifnotheadornothead.next:8returnTrue910m =011start =head12whilestart:13start =start.next14m += 11516n = 117stack =[]18start =head19ifm%2 ==0:20whilen <= ...
234. Palindrome Linked List 難度 easy 個人範例程式碼 – Deque (比較進階),空間 O(N) # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def isPalindrome(self, head: Optional[ListNode]...
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
然后我们从慢指针的下一个开始,把后面的链表都反转(Reverse Linked List),然后我们再从头和从尾同时向中间前进,就可以判断该链表是不是回文了。 代码 public class Solution { public boolean isPalindrome(ListNode head) { if(head == null || head.next == null) return true;...
Leetcode-Easy 234. Palindrome Linked List 描述: 判断一个单链表是否左右对称 代码语言: 代码运行次数: # Definitionforsingly-linked list.#classListNode:# def__init__(self,x):# self.val=x # self.next=NoneclassSolution(object):defisPalindrome(self,head):""":type head:ListNode:rtype:bool"""...
class Solution { public: bool isPalindrome(ListNode* head) { if(head == NULL || head->next == NULL) return true; stack<int> st; ListNode *tmp = head; while(tmp) { st.push(tmp->val); tmp = tmp->next; } while(head)
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 ...
publicclassSolution{ publicbooleanisPalindrome(ListNodehead){ ListNodeend=head; ListNodemid=head; while(end!=null&&end.next!=null){ end=end.next.next; mid=mid.next; } if(end!=null)//in case of odd list mid=mid.next; mid=reverseList(mid); ...
Solution Analysis: First convert the linked list into an array, and then judge whether the array is a palindrome array Loop this array to determine whether the values after moving the same distance from the start and end positions are the same. If they are all the same, the result is a ...