Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 解决这个问题需要很巧妙的思路,一般我们会想到观察链表中是否出现过重复出现的节点.但是这样的空间的复杂度是O(n),比如用set来存储出现过的节点,然后看下一个节点是否存在于set中. 一个很巧妙的解决办法是设置两个指针,一个快指针和一个慢指针.快指针每次移动两步,慢指针每次移动一…

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 解法一: 使用unordered_map记录当前节点是否被访问过,如访问过说明有环,如到达尾部说明无环. /** * Definition for singly-linked list. * struct ListNode { * int va…

Problem link: http://oj.leetcode.com/problems/linked-list-cycle/ We set two pointers: the faster pointer goes two steps each iteration, and the slower one goes one step. If the two pointers meet some time, it follows that there is a loop; otherwise,…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? (二)解题 本题大意:给定一个链表,判断链表里面是否成环.不能用辅助空间. 解题思路:利用快…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 双指针 保存已经走过的路径 日期 [LeetCode] 题目地址:https://leetcode.com/problems/linked-list-cycle/ Total Accepted: 102417 Total Submissions: 277130 Difficulty: Easy 题目描述 Given a linked list, de…

Problem link: http://oj.leetcode.com/problems/linked-list-cycle-ii/ The solution has two step: Detecting the loop using faster/slower pointers. Finding the begining of the loop. After two pointers meet in step 1, keep one pointer and set anther to th…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up:Can you solve it without using extra space? 这个题还是蛮考验数学推理的,不过在前一个题的基础上还是能推出结果的.这是英文一段解释,非常有帮助. First Step: A…

题目: Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 提示: 首先,题目中要求'without using extra space',指的是空间复杂度必须控制在O(1)内. 因此可以创建两个变量,先同时指向head,然后每一轮循环中,令其中一个变量沿链表向前“走”两步,另一个走“一步”,这样的话每一个循环后他们两者的距离差会…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up:Can you solve it without using extra space? 和前面那题类似,只不过这里要找到的是环开始的节点,看下面这张图(图是盗的): 当fast以及slow指针在z处相遇的时候,可以…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 判断一个链表是否存在环,维护快慢指针就可以,如果有环那么快指针一定会追上慢指针,代码如下: class Solution { public: bool hasCycle(ListNode *head) { ListNode * slow, * fast; slow = fast…