Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 这道题是快慢指针的经典应用.只需要设两个指针,一个每次走一步的慢指针和一个每次走两步的快指针,如果链表里有环的话,两个指针最终肯定会相遇.实在是太巧妙了,要是我肯定想不出来.代码如下: C++ 解法: class Solution { public: bool hasCycle…

题目: Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle…

这道题是LeetCode里的第141道题. 题目要求: 给定一个链表,判断链表中是否有环. 进阶: 你能否不使用额外空间解决此题? 简单题,但是还是得学一下这道题的做法,这道题是用双指针一个fast,一个slow.fast每一步前进两个节点,slow前进一个节点.判断fast和slow是否相等或者为空就行了. 贴个代码: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 判断链表中是否有环,不能用额外的空间,可以使用快慢指针,慢指针一次走一步,快指针一次走两步,若是有环则快慢指针会相遇,若是fast->next==NULL则没有环. 值得注意的是:在链表的题中,快慢指针的使用频率还是很高,值得注意. /** * Definition for si…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? Hide Tags Linked List Two Pointers        开始犯二了一点,使用了node 的val 作为判断,其实是根据内存判断.找出链表环的起始位置,这个画一下慢慢找下规律…

给定一个链表,判断链表中否有环.补充:你是否可以不用额外空间解决此题?详见:https://leetcode.com/problems/linked-list-cycle/description/ Java实现: /** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } *…

题意:给一个单链表,判断其是否出现环! 思路:搞两个指针,每次,一个走两步,另一个走一步.若有环,他们会相遇,若无环,走两步的指针必定会先遇到NULL. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool has…

2.6 Given a circular linked list, implement an algorithm which returns the node at the beginning of the loop.DEFINITIONCircular linked list: A (corrupt) linked list in which a node's next pointer points to an earlier node, so as to make a loop in the…

Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 和问题一Linked List Cycle几乎一样.如果用我的之前的解法的话,可以很小修改就可以实现这道算法了.但是如果问题一用优化了的解法的话,那么就不适…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 这个求单链表中的环的起始点是之前那个判断单链表中是否有环的延伸,可参见我之前的一篇文章 (http://www.cnblogs.com/grandyang/p/4137187.html). 还是要设…

Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? 141. Linked List Cycle 的拓展,这题要返回环开始的节点,如果没有环返回null. 解法:双指针,还是用快慢两个指针,相遇时记下节点.参考:willduan的博客 Java: pu…

题目要求 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? 如何判断一个单链表中有环? Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle…

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull. Follow up:Can you solve it without using extra space? 题意:给定链表,若是有环,则返回环开始的节点,没有则返回NULL 思路:题目分两步走,第一.判断是否有环,第二若是有,找到环的起始点.关于第一点,可以参考之前的博客 Linked list cycle.…

Given a linked list, determine if it has a cycle in it. ExampleGiven -21->10->4->5, tail connects to node index 1, return true Challenge Follow up:Can you solve it without using extra space? LeetCode上的原题,请参见我之前的博客Linked List Cycle. class Solution…

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? SOLUTION 1: 经典快慢指针问题.如果存在环,fast, slow必然会相遇.就像2个速度不一样的人在环形跑道赛跑,总有一个时间他们会相遇. 如果fast到达了null,就是没有环,可以返回false. package Algorith…

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? Linked List Two Pointers     ''' Created on Nov 13, 2014 @author: ScottGu<gu.kai.66@gmail.com, 150316990@qq.com> ''' # De…

Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? SOLUTION 1: 1. 先用快慢指针判断是不是存在环. 2. 再把slow放回Start处,一起移动,直到二个节点相遇,就是交点.…

目录 141_环形链表 描述 解法一:哈希表 思路 Java 实现 Python 实现 解法二:双指针(龟兔算法) 思路 Java 实现 Python 实现 141_环形链表 描述 给定一个链表,判断链表中是否有环. 进阶: 你能否不使用额外空间解决此题? 解法一:哈希表 思路 判断一个链表是否包含环,可以转化为判断是否有一个节点之前已经出现过.非常自然的一个想法就是:遍历链表的每个节点,用一个哈希表记录每个节点的引用(或内存地址):如果能够遍历到空节点,则此时已经遍历到链表的尾部,返回 fal…

141. 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? 利用快慢指针,如果相遇则证明有环 注意边界条件: 如果只有一个node. public class Solution { public boolean hasCycle(ListNode head) { if(head==null |…

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? 错误解法.因为Cycle可能出现在Link的中间的,所以需要检查其中间Nodes. bool hasCycle(ListNode *head) { // IMPORTANT: Please reset any member data you…

141题: 首先,先看141题,这个题是比较初级也是比较经典的环形链表题: 给定一个链表,判断链表中是否有环. 进阶:你能否不使用额外空间解决此题? 那么,什么是有环的链表呢: 这个就是有环的链表 题设中说,能否不使用额外空间解决此题,说明,最普通的方法就是用其他的数据结构解决这个问题,那么我们想把链表结点依次遍历放入哈希表中,然后遇到重复的结点说明是有环的,这就是利用了其他的额外空间来解决这个问题,空间复杂度为O(n),现在我们直接说进阶的思路: 思路 如果,在环形的跑道上跑步的话,跑的快和跑…

Question : Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? Anaylsis : 首先,比较直观的是,先使用Linked List Cycle I的办法,判断是否有cycle.如果有,则从头遍历节点,对于每一个节点,查询是否在环里面,是…

Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer poswhich represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in t…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 题意: 给定一个链表,判断是否有环 思路: 快慢指针 若有环,则快慢指针一定会在某个节点相遇(此处省略证明) 代码: public class Solution { public boolean hasCycle(ListNode head) { ListNode fast =…

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…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 题目标签:Linked List 题目给了我们一个 Linked List,让我们判断它是否循环. 利用快,慢指针,快指针一次走2步,慢指针一次走1步,如果循环,快慢指针一定会相遇. Java Solution: Runtime beats 98.15% 完成日期:06/09/2…

题意:给一个单链表,若其有环,返回环的开始处指针,若无环返回NULL. 思路: (1)依然用两个指针的追赶来判断是否有环.在确定有环了之后,指针1跑的路程是指针2的一半,而且他们曾经跑过一段重叠的路(即1跑过,2也跑过),就是那段(环开始处,相遇处),那么指针2开始到环开始处的距离与head到指针相遇处是等长的喔~,那么再跑一次每次一步的就必定会相遇啦.画个图图好方便看~ (2)其实还有另一个直观的思路,就是指针1和2相遇后,p指向他们的next,在他们相遇处的next给置空,再跑一遍那个“找两…

题意:判断链表是否有环. 分析:快慢指针. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool hasCycle(ListNode *head) { ListNode* fast = head; ListNode…

问题描述如下: Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 从问题来看,如果可以充分利用额外空间的话,这个题目是不难的,然而题目提出了一个要求,能否在不使用任何额外空间的情况下解决这个问题. 通过反复思考,我觉得这题类似于追击问题,可以用一个…