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
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

跑两步的追跑一步的~

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool hasCycle(ListNode *head) {
        ListNode* fa=head;
        ListNode* sl=head;
        while(fa&&fa->next){
            fa=fa->next;
            fa=fa->next;
            sl = sl->next;
            if(sl==fa){return true;}
        }return false;
    }
};

最快的老哥和我差在哪里？ 

if(head == nullptr || head->next == nullptr)

            return false;

加了这一个判断就往前走了不少20ms变成8ms

 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    
    bool hasCycle(ListNode *head) {
        if(head == nullptr || head->next == nullptr)
            return false;
        
        ListNode* slow = head->next;
        ListNode* fast = slow->next;
  
        
        while(slow != fast){
            if(slow == nullptr || fast == nullptr)
                return false;
            slow = slow->next;
            fast = fast->next;
            if(fast != nullptr)
                fast = fast->next;
        }
        return true;
    }
};