leetcode 141 Linked List Cycle Hash fast and slow pointer

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

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?


Ac Code: (Hash)

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     bool hasCycle(ListNode *head) {
12         unordered_set<ListNode*> visit;
13         while(head)
14         {
15             if(visit.count(head)!=0) return true;
16             visit.insert(head);
17             head = head->next;
18         }
19         return false;
20     }
21 };

 Fast and Slow Pointer

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     bool hasCycle(ListNode *head) {
12         ListNode *slow = head;
13         ListNode *fast = head;
14         while(fast){
15             if(!fast->next) return false;
16             fast = fast->next->next;
17             slow = slow->next;
18             if(slow == fast) return true;
19         }
20         return false;
21     }
22 };

 

posted on 2019-07-13 19:18  Algorithm_Ethusiast  阅读(124)  评论(0编辑  收藏  举报

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