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 the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

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(ListNode *head) {
ListNode *slow = head, *fast = head;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
if (slow == fast) return true;
}
return false;
}
};

Java 解法:

public class Solution {
public boolean hasCycle(ListNode head) {
ListNode slow = head, fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) return true;
}
return false;
}
}

Github 同步地址:

https://github.com/grandyang/leetcode/issues/141

类似题目:

Linked List Cycle II

Happy Number

参考资料:

https://leetcode.com/problems/linked-list-cycle/

https://leetcode.com/problems/linked-list-cycle/discuss/44489/O(1)-Space-Solution

LeetCode All in One 题目讲解汇总(持续更新中...)

[LeetCode] 141. Linked List Cycle 单链表中的环的更多相关文章

  1. [CareerCup] 2.6 Linked List Cycle 单链表中的环

    2.6 Given a circular linked list, implement an algorithm which returns the node at the beginning of ...

  2. [LeetCode] 142. Linked List Cycle II 链表中的环 II

    Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Foll ...

  3. [LeetCode] Linked List Cycle 单链表中的环

    Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using ex ...

  4. LeetCode 141. Linked List Cycle 判断链表是否有环 C++/Java

    Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked lis ...

  5. [LintCode] Linked List Cycle 单链表中的环

    Given a linked list, determine if it has a cycle in it. ExampleGiven -21->10->4->5, tail co ...

  6. LeetCode 141. Linked List Cycle(判断链表是否有环)

    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; * List ...

  7. [leetcode]141. Linked List Cycle判断链表是否有环

    Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using ext ...

  8. 141. Linked List Cycle(判断链表是否有环)

    141. Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you sol ...

  9. LeetCode 141. Linked List Cycle环形链表 (C++)

    题目: Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked ...

随机推荐

  1. 移动端js触摸touch详解(附带案例源码)

    移动端触摸滑动原理详解案例,实现过程通过添加DOM标签的触摸事件监听,并计算触摸距离,通过距离坐标计算触摸角度,最后通过触摸角度去判断往哪个方向触摸的. 触摸的事件列表 触摸的4个事件: touchs ...

  2. mysql8 安装

    准备工作: 首先安装这些依赖 yum install -y flex yum install gcc gcc-c++ cmake  ncurses ncurses-devel bison libaio ...

  3. Spring AOP中使用@Aspect注解 面向切面实现日志横切功能详解

    引言: AOP为Aspect Oriented Programming的缩写,意为:面向切面编程,通过预编译方式和运行期动态代理实现程序功能的统一维护的一种技术.AOP是OOP的延续,是软件开发中的一 ...

  4. vuex 源码分析(一) 使用方法和代码结构

    Vuex 是一个专为 Vue.js 应用程序开发的状态管理模式,它采用集中式存储管理应用的所有组件的状态,注意:使用前需要先加载vue文件才可以使用(在node.js下需要使用Vue.use(Vuex ...

  5. LaTex语法

    排版数学公式是TeX系统设计的初衷,它在LaTeX中占有特殊地位,也是LaTeX最为人所称道的功能之一.基于对MathType排版效果的不满意,以及对公式进行检索的需求,我们使用LaTeX输入数学公式 ...

  6. 记录自己运行eShopOnContainers过程中遇到的坑

    由于各种各样的问题,依照官方文档运行eShopOnContainers项目遇到了好多莫名其妙的错误. 好在最后都解决了,在此记录,以防自己以后再遇到,也为遇到同样问题的同学提供参考. 参考的官方文档 ...

  7. Activex在没有电子秤api的情况下获取串口数据

    大二做B/S架构的项目使用了安衡电子秤CHS-D+R和一款扫码枪,两个设备的串口使用一样,这款电子秤是相当的坑,没有开发的api,无奈只能自己开发Activex了,在B/S架构中进行引用Activex ...

  8. //某父元素(.class)底下相同class的第二的取值

    //某父元素(.class)底下相同class的第二的取值 var v = $('.cell-right').find(".startime").eq(1).val();

  9. pytest-Mark数据驱动

    数据驱动 import pytest @pytest.mark.parametrize(("a", "b", "expected"), [ ...

  10. Ansible Jinja2 模板

    1.jinja2渲染NginxProxy配置文件 jinja2 房屋建筑设计固定的? jinja2模板与Ansible关系 Ansible如何使用jinja2模板 template模块 拷贝文件? t ...