引入 快慢指针经常用于链表(linked list)中环(Cycle)相关的问题.LeetCode中对应题目分别是: 141. Linked List Cycle 判断linked list中是否有环 142. Linked List Cycle II 找到环的起始节点(entry node)位置. 简介 快指针(fast pointer)和慢指针(slow pointer)都从链表的head出发. slow pointer每次移动一格,而快指针每次移动两格. 如果快慢指针能相遇,则证明链表中有…

判断链表有环,环的入口结点,环的长度 1.判断有环: 快慢指针,一个移动一次,一个移动两次 2.环的入口结点: 相遇的结点不一定是入口节点,所以y表示入口节点到相遇节点的距离 n是环的个数 w + n + y = 2 (w + y) 经过化简,我们可以得到:w  = n - y; https://www.cnblogs.com/zhuzhenwei918/p/7491892.html 3.环的长度: 从入口结点或者相遇的结点移动到下一次再碰到这个结点计数 https://blog.csdn.ne…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 给定一个链表,判断是否有环存在.Follow up: 不使用额外空间. 解法:双指针,一个慢指针每次走1步,一个快指针每次走2步的,如果有环的话,两个指针肯定会相遇. Java: public class Solution { public boolean hasCycle(Li…

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, 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…

Problem describe:https://leetcode.com/problems/linked-list-cycle/ Given a linked list, determine if it has a cycle in it. To represent a cycle -indexed) , then there is no cycle in the linked list. Example : Input: head = [,,,-], pos = Output: true E…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 题意: 判断一个链表是否有环. 解决方案: 双指针,快指针每次走两步,慢指针每次走一步, 如果有环,快指针和慢指针会在环内相遇,fast == slow,这时候返回true. 如果没有环,返回false. /** * Definition for singly-linked li…

Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 解题思路: 由于不让用extra space,所以用一个快指针和一个慢指针,快指针一次移动两步,慢指针一次移动一步,只要快指针赶上慢指针证明纯在loop,JAVA实现如下: public boolean hasCycle(ListNode head) { ListNode fa…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 判断是否成环 1.利用set,很简单,但是题目中说不要用额外的空间. /** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x)…

题目描述: Given a linked list, determine if it has a cycle in it. 解题思路: 快的指针和慢的指针 代码如下: # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def hasCycle(self, h…