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...
n= n / 2;while(n-- > 0) tmpHead =tmpHead.next; ListNode reverse= reverseList(tmpHead);//反转while(reverse !=null){//比较反转后的链表,与正序链表if(head.val != reverse.val)returnfalse; head=head.next; reverse=reverse.next; }returntrue; }privateListNode reverseList(ListNode head){//...
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def isPalindrome(self, head): """ :type head: ListNode :rtype: bool """ node_list=[] while head: node_list.append(head) head=head.ne...
Could you do it in O(n) time and O(1) space? 思路:快慢指针找到中点,同时修改前一半链表的.next,然后从中间开始向链表两端比较。 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class ...
// this change would reflect in the parent recursive calls left = left->next; // In order for linked list to be palindrome, the character at the left // node should match with the character at the right node return (prevLeft->data == right->data); ...
Write a C program to check if a singly linked list is a palindrome or not. Sample Solution: C Code: #include<stdio.h>#include<stdlib.h>#include<stdbool.h>// Node structure for the linked liststructNode{intdata;structNode*next;};// Function to create a new nodestructNode*newNode(int...
Palindrome Linked List 文章作者:Tyan 博客:noahsnail.com | CSDN | 简书 1. Description 2. Solution Reference https://leetcode.com/problems/palindrome-linked-list/description/ Leetcode 36. Valid Sudoku 文章作者:Tyan 博客:noahsnail.com | CSDN | 简书 1. Description 2. Solution Reference https:...
代表list是本次的结果 if (pos == c.length) // 添加每一次的结果 res.add(new ArrayList<>(list)); // 到了此字符串的pos下标了,还有c 这个剩余字符串等待处理 for (int i = pos; i < c.length; i++) { //如果是回文,接着往下递归 if (isPal(c, pos, i)){ // String(c, pos, i ...
234. Palindrome Linked List 原题链接:https://leetcode.com/problems/palindrome-linked-list/ 我自己的想法是把这个链表复制一份然后倒过来,对比一下两个链表是否一样。想法简单但是操作起来非常复杂···链表的复制和比较都用自定义func来实现的。 顺便总结一下copy.copy和copy.deepcopy的区别...
234 palindrome linked list 1,找到中间节点然后反转后半部, 2,从头到尾依次比较每个节点的值 当链表个数是偶数个数的时候很好理解,当时奇数的时候有一个比较tricky的地方 1->2->3->NULL 反转后半部之后为3->2->NULL. 前半部1->还是指向2 1->2->NULL....