题目

level：easy

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?

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/linked-list-cycle

解题思路

想象有两个指针，一个指针每次只往后移动一个节点，另一个指针每次往后移动两个节点，若节点能形成环，则这两个指针在移动过程中必定会重合；

1. 声明两个指针，一个指针每次移动一个节点，另一个指针每次移动两个节点；
2. 当遇到NULL时，必定不成环；若遍历过程中两节点相遇，则必定成环。

算法实现（C++）

/**1. Definition for singly-linked list.2. struct ListNode {3.     int val;4.     ListNode *next;5.     ListNode(int x) : val(x), next(NULL) {}6. };*/
class Solution {
public:bool hasCycle(ListNode *head) {if (!head)return false;ListNode* oneNode = head;//该指针每次只往后移动一个节点ListNode* twoNode = head->next;//该指针每次往后移动两个节点while (oneNode && twoNode && twoNode->next){if (oneNode == twoNode)return true;oneNode = oneNode->next;twoNode = twoNode->next->next;}//程序运行至此，则说明退出了while循环，说明上述节点有NULL存在，则必定不成环return false;}
};

复杂度分析

1. 时间复杂度：O(n)，当链表不成环时，节点遍历到NULL则结束；当链表成环时，则在成环节点的有限圈数内两指针必定相遇；
2. 空间复杂度：O(1)，只需有限的两个指针即可。