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  1. Cycle Detection
  2. Discussions

Cycle Detection

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32 Discussions

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  • misha_y1404
     + 0 comments

    Only python works for me...

    Floyd cycle approach

    def has_cycle(head):
    slow = head
    fast = head
    
    while fast is not None and fast.next is not None:
        slow = slow.next
        fast = fast.next.next
        
        if (slow == fast):
            return 1
    
    return 0

  • vadym_usatiuk
     + 1 comment

    static boolean hasCycle(SinglyLinkedListNode head) { if (head == null || head.next == null) { return false; } SinglyLinkedListNode slow = head; SinglyLinkedListNode fast = head.next; while (fast != null && fast.next != null) { if (slow == fast) { return true; } slow = slow.next; fast = fast.next.next; } return false; }

  • balla_kapo
     + 0 comments
    def has_cycle(head):
    visited = set()
    while n:=head.next:
        if n in visited:
            return 1
        else:
            visited.add(n)
            head = n
    return 0

  • bhaweshwebmaster
     + 0 comments

    C++

    bool has_cycle(SinglyLinkedListNode* head) {
    unordered_set<SinglyLinkedListNode*> address; // a set of SLL num_set
    
    // for every node until the end of the SLL    
    while (head){        
        if (address.find(head) != address.end()) // if found the head in address
            return 1; // return 1 and terminate
        else 
            address.insert(head); // if did not find the head in address, then insert head as another element in address
            
        head = head->next; // move to the next node in the SLL
    }
    return 0; // if we got here then return zero
}

  • guysneh
     + 0 comments

    C# Solution:

    static bool hasCycle(SinglyLinkedListNode head) {

        List<SinglyLinkedListNode> list = new List<SinglyLinkedListNode>();
        while(head!=null) 
        {
            if (list.Contains(head)) 
            {
                return true;
            } 
            else 
            {
                list.Add(head);
                head = head.next;
            }
        }
        return false;
    }