力扣算法——141LinkedListCycel【E】

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?

Accepted

Solution:

　　使用快慢指针，若两个指针会重合，那么就有环，否则没有

 class Solution {

 public:

     bool hasCycle(ListNode *head) {

         if (head == nullptr || head->next == nullptr)return false;

         ListNode *slow, *fast;

         slow = fast = head;

         while (fast && fast->next)

         {

             slow = slow->next;

             fast = fast->next->next;

             if (fast == slow)

                 return true;

         }

         return false;

     }

 };

巴特西

力扣算法——141LinkedListCycel【E】

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