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, return the node where the cycle begins. If there is no cycle, return null. 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.…

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…

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…

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.…

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…

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…

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…

一.题目大意 https://leetcode.cn/problems/linked-list-cycle-ii/ 给定一个链表的头节点  head ,返回链表开始入环的第一个节点. 如果链表无环,则返回 null. 如果链表中有某个节点,可以通过连续跟踪 next 指针再次到达,则链表中存在环. 为了表示给定链表中的环,评测系统内部使用整数 pos 来表示链表尾连接到链表中的位置(索引从 0 开始).如果 pos 是 -1,则在该链表中没有环.注意:pos 不作为参数进行传递,仅仅是为了标识链…

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…