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 };