Given a singly linked list, determine if it is a palindrome. Example 1: Input: 1->2 Output: false Example 2: Input: 1->2->2->1 Output: true Follow up:Could you do it in O(n) time and O(1) space? Sloving with O(3n) -> O(n) Idea: reversing th…

http://www.geeksforgeeks.org/reverse-a-list-in-groups-of-given-size/ #include <iostream> #include <vector> #include <algorithm> #include <queue> #include <stack> #include <string> #include <fstream> #include <m…

9 - Palindrome Number Determine whether an integer is a palindrome. Do this without extra space. Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using extra…

234. Palindrome Linked List 1. 使用快慢指针找中点的原理是fast和slow两个指针,每次快指针走两步,慢指针走一步,等快指针走完时,慢指针的位置就是中点.如果是偶数个数,正好是一半一半,如果是奇数个数,慢指针正好在中间位置,判断回文的时候不需要比较该位置数据. 注意好好理解快慢指针的算法原理及应用. 2. 每次慢指针走一步,都把值存入栈中,等到达中点时,链表的前半段都存入栈中了,由于栈的后进先出的性质,就可以和后半段链表按照回文对应的顺序比较. solution…

234. Palindrome Linked List[easy] 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? 解法一: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * Lis…

Question 234. Palindrome Linked List Solution 题目大意:给一个链表,判断是该链表中的元素组成的串是否回文 思路:遍历链表添加到一个list中,再遍历list的一半判断对称元素是否相等,注意一点list中的元素是Integer在做比较的时候要用equals,因为int在[-128,127)之间在内存中是存储在运行时常量池中,超出这个范围作为对象存储在堆中,==比较的是字面值和对象地址 Java实现: public boolean isPalindrom…

一.Reverse Linked List  (M) Reverse Linked List II (M) Binary Tree Upside Down (E) Palindrome Linked List /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class…

Total Accepted: 48614 Total Submissions: 185356 Difficulty: Hard Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should rem…

1. Reverse Linked List 题目链接 题目要求: Reverse a singly linked list. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 初想用递归来实现应该会挺好的,但最终运行时间有点久,达到72ms,虽然没超时.具体程序如下: /** * Definition for singly-linked list. *…

It helps to understands how recursive calls works. function Node(val) { return { val, next: null }; } function LinkedList() { return { head: null, tail: null, add(val) { const node = new Node(val); if (!this.head) { this.head = node; this.tail = node…