isPalindrome(); } return0; } This article tried to discuss different approaches for finding palindrome linked list by which topics like stack, recursion and linked list are improved. Having knowledge about the
Problem: Implement a function to check if a singly linked list is a palindrome. 思路: 最简单的方法是 Reverse and compare. 另外一种非常经典的办法是用 Recursive 的思路,把一个list看成这种形式: 0 ( 1 ( 2 ( 3 ) 2 ) 1 ) 0 0 ( 1 ( 2 ( 3 3 ) 2 ) 1 ) 0 CC150里面给出的Code...
这就是个palindrome,开头的1和结尾的1对应,第二位的2和倒数第二位的2对应 我查到一个不错的方法,大家可以参考一下,思路很好,利用了栈FIFO的性质 public static boolean isPalindrome(Node head) { Node slwptr=head; Node fstptr=head; Stack<Integer> s1=new Stack<Integer>(); while(fstptr!=null && ...
public static boolean checkPalindrome (Node head) { // Find middle node using slow and fast pointer Node middleNode=findMiddleNode(head); // we got head of second part Node secondHead=middleNode.next; // It is end of first part of linked list middleNode.next=null; // get reversed lin...
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...
{ next = current->next; current->next = prev; prev = current; current = next; } return prev; } // Function to check if a linked list is a palindrome bool isPalindrome(struct Node* head) { struct Node* slow = head; struct Node* fast = head; struct Node* prev_slow = head; ...
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Learn how to check if a string is a palindrome in JavaScript while considering punctuation. This guide provides clear examples and explanations.
Exception Handling Recursion Sorting and Searching Linked Lists Stacks and Queues Sets and Maps Generic Classes Bug Report Form Your name: Your email address: Problem description: To protect against spam robots, please answer this simple math problem: *=...
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