http://www.geeksforgeeks.org/function-to-check-if-a-singly-linked-list-is-palindrome/ 这里的reverse可以reverse整个list,这样空间需求就是O(n),不如这个网页写的O(1)的方法 1#include <iostream>2#include <vector>3#include <algorithm>4#include <queue>5#include <stack>6#include <string>7#include...
依次比较p1 与 p2 节点值是否相等。 1//This code would destroy the original structure of the linked list.2//If you do not want to destroy the structure, you can reserve the second part back.3publicclassSolution {4/**5*@paramhead a ListNode6*@returna boolean7*/8publicbooleanisPalindrome(...
Implement a function to check if a linked list is a palindrome. Example Given1->2->1, return true. 分析: 比较容易的方法是把所有的数存在ArrayList里,然后查看是否是palindrome,但是空间复杂度为O(n)。 还有就是先找到list的中点,然后把后半段reverse,然后再进行比较。但是这种方法不容易想到。 1 /**...
class Solution { public boolean isPalindrome(ListNode head) { if (head == null || head.next == null) return true; ListNode fast = head; ListNode newHead = null; while (fast != null) { if (fast.next == null) { //有奇数个节点的时候,会在最后一步走这个if分支,head往前走一步,意思...
什么是单向链表(singly linked list)? 1单向链表是一种线性的数据结构,但是其元素的存储并不是连续的,而是通过指针(pointer)来指向下一个元素。 2单向链表的每一个元素我们称之为node,每一个node包含两个字段:data(用于存储数据),link(指针来指向下一个元素)。
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。 /** * Definition for singly-linked list. ...
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
determine whether a phrase is a palindrome How to save a linked list The differences between using linked lists and arrays for storing a list of items How to represent a linked list using arrays How to merge two sorted linked lists The concept of a circular list and a doubly linked listAdv...
Given the head of a singly linked list, return true if it is a palindrome. Example 1: Input: head = [1,2,2,1] Output: true Example 2: Input: head = [1,2] Output: false Constraints: The number of nodes in the list is in the range[1, 105]. ...
classSolution{publicListNodereverseList(ListNodehead){ListNodenext;ListNodecurr=head;ListNodenewlist=null;while(curr!=null){next=curr.next;curr.next=newlist;newlist=curr;curr=next;}returnnewlist;}publicbooleanisPalindrome(ListNodehead){// Find middle nodeintsize=0,count=0;ListNodecurr=head,middle_no...