Given a singly linked list, determine if it is a palindrome. 推断一个链表是不是回文的,一个比較简单的办法是把链表每一个结点的值存在vector里.然后首尾比較.时间复杂度O(n).空间复杂度O(n). /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x),…

2.7 Implement a function to check if a linked list is a palindrome. LeetCode上的原题,参见我之前的博客Palindrome Linked List 回文链表.…

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…

Palindrome Linked List 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: 一次遍历,装入vector,然后再一次遍历判断回文. 时间复杂度O(n),空间复杂度O(n) /** * Definition for singly-linked list. * struct ListNode…

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…

234. Palindrome Linked List Easy 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? package leetcod…

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…

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? 这道题让我们判断一个链表是否为回文链表,LeetCode中关于回文串的题共有六道,除了这道,其他的五道为Palindrome Number 验证回文数字,Validate Palindrome 验证回文字符串,Palindrome Partitioning 拆分回文…

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? 题目意思: 给定一个单链表,判断它是不是回文串 进一步思考: 你可以在O(n)时间复杂度和O(1)空间复杂度完成吗? 解题思路: 方法一:通过反转链表实现 (1)使用快慢指针寻找链表中点 (2)将链表的后半部分就地逆置 (3)比较前后两半的元素是否一致 (4)恢复原始…

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? /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} *…

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? 思想:转置后半段链表节点,然后比较前半段和后半段节点的值是否相等. 代码如下: /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next;…

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? 分析:题意为判断单链表是否为回文的. 思路:首先想到的是 遍历一次单链表,将其元素装入vector,然后进行第二次遍历比较来判断回文. /** * Definition for singly-linked list. * struct ListNode { * int…

Given a singly linked list, determine if it is a palindrome. 思路: 用快慢指针找到链表中点,反转后半部分链表,然后与前半部分进行匹配,随后将链表恢复原状(本题没有这个要求,具体情况具体对待). C++: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x),…

题目: 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? 链接: http://leetcode.com/problems/palindrome-linked-list/ 题解: 判断链表是否是Palindrome. 我们分三步解,先用快慢指针找中点,接下来reverse中点及中点后部,最后逐节点对比值. Time…

题目描述: 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? 解题思路: 使用O(n)的时间复杂度及O(1)的时间复杂度表明顺序遍历链表以及不能够开辟跟链表相当的空间,这样可以找出中间的位置,然后将后半部分的链表反转,然后跟前半部分的链表逐个位置比对. 代码如下: /** * Definition for singl…

Given a singly linked list, determine if it is a palindrome. 思路: 回文结构从后向前遍历与从前向后遍历的结果是相同的,可以利用一个栈的结构,将出栈元素与正向移动的指针指向元素比较,即可判断. 解法: /* public class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } */ import java.util.Stack; public cla…

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? 问题:给定一个单向列表结构,判断它是不是回文的. 补充:是否可以在 O(n) 时间,O(1) 额外空间下完成? 解题思路: 对于数组,判断是否是回文很好办,只需要用两个指针,从两端往中间扫一下就可以判定. 对于单向列表,首先想到的是,将列表复制一份到数组中,然后用上面…

Question 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? Solution 这一题思路并不难.要满足follow-up的要求,我们用到了快慢指针. 1. 用快慢指针得到前后两半list,这里有个技巧是quick先判断有无next,slow再走.这样就保证slow永远指向后半部分的前一个结点 2. Rever…

Total Accepted: 29652 Total Submissions: 117516 Difficulty: 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?  (E) Palindrome Number (E) Valid Palindrome (E) Reverse Linked List /…

题目: Given a singly linked list, determine if it is a palindrome. 判断一个单链表是不是回文 思路: 1.遍历整个链表,将链表每个节点的值记录在数组中,再判断数组是不是一个回文数组,时间复杂度为O(n),但空间复杂度也为O(n),不满足空间复杂度要求. 2.利用栈先进后出的性质,将链表前半段压入栈中,再逐个弹出与链表后半段比较.时间复杂度O(n),但仍然需要n/2的栈空间,空间复杂度为O(n). 3.反转链表法,将链表后半段原地翻转,…

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? Subscribe to see which companies asked this question   思路: 1.利用快慢两个指针找到中间结点 2.从中间结点下一个结点开始逆置链表,得到新的链表 3.新的链表与原来链表依次比较结点,如有不同,return fa…

[题目描述] Implement a function to check if a linked list is a palindrome. 设计一种方式检查一个链表是否为回文链表. [题目链接] www.lintcode.com/en/problem/palindrome-linked-list/ [题目解析] 假设一个链表是回文,那么把链表分割为1-n/2,n/2+1-n两个部分,这两个部分肯定是相同的(把第二部分顺序逆转过来或者逆转第一部分).所以如果我们能把任何一个部分的链表顺序逆转过来…

题目: Given a singly linked list, determine if it is a palindrome. Follow up: 思路: 题意:判断一个链表是不是回文 利用两个指针,一个快fast,一个慢slow,slow = head,fast = head,每次fast跳两格,slow一格,直到fast.next.next == null结束,slow位于中间,把slow.next往后的元素倒叙.prev = head,比较head和end的val 代码: /** *…

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? 请判断一个链表是否为回文链表. 示例 1: 输入: 1->2 输出: false 示例 2…

Description: 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? 判断一个单向链表是不是回文. 思路: 最简单的思路就是,用一个遍历一遍,存下来,再遍历反着做比较.时间复杂度O(n),空间复杂度O(n).竟然能过...... /** * Definition for singly-linked list.…

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: 1. 利用快慢指针找到链表中点 2.从中点开始反转链表,判断是否相等. class Solution { public boolean isPalindrome(ListNode head) { ListNode faster= head ,slower= h…

Total Accepted: 40445 Total Submissions: 148124 Difficulty: 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? 分析: 临时没想到空间为O(1)的方法,直白的模拟思想,遍历一遍记录各节点的元素 再对数组前尾比較就可以. /** * Definiti…

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? /** * Definition for singly-linked list. * publi…

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? 判断一个链表是否为回文链表,有很多方法都可以进行判断,可以先遍历链表然后将值全部存储到vector之中,也可以遍历链表然后将list中的值都压倒堆栈中再和List进行比较, 但是想要达到题目上面的 时间复杂度为O(n),而空间复杂度为O(1)还要用到递归: 首先是不用…