C++实现LeetCode(141.单链表中的环)

[LeetCode] 141. Linked List Cycle 单链表中的环

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.

C++实现LeetCode(141.单链表中的环)

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.

C++实现LeetCode(141.单链表中的环)

Example 3:

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

C++实现LeetCode(141.单链表中的环)

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;

									    }

									}