Cycle Detection
Problem Statement :
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refers 1 -> 2 -> 3 -> 1 -> NULL There is a cycle where node 3 points back to node 1, so return 1. Function Description Complete the has_cycle function in the editor below. It has the following parameter: SinglyLinkedListNode pointer head: a reference to the head of the list Returns int: 1 if there is a cycle or 0 if there is not Note: If the list is empty, head will be null. Input Format The code stub reads from stdin and passes the appropriate argument to your function. The custom test cases format will not be described for this question due to its complexity. Expand the section for the main function and review the code if you would like to figure out how to create a custom case.
Solution :
Solution in C :
In C++ :
/*
Detect loop in a linked list
List could be empty also
Node is defined as
struct Node
{
int data;
struct Node *next;
}
*/
int HasCycle(Node* head)
{
// Complete this function
// Do not write the main method
Node *slowp = head, *fastp = head;
while (slowp && fastp && fastp->next)
{
slowp = slowp->next;
fastp = fastp->next->next;
if (slowp == fastp)
{
return 1;
}
}
return 0;
}
In Java :
/*
Insert Node at the end of a linked list
head pointer input could be NULL as well for empty list
Node is defined as
class Node {
int data;
Node next;
}
*/
int HasCycle(Node head) {
if(head==null)
return 0;
if(head.next==head)
return 1;
Node slow=head;
Node fast=head;
while(fast!=null)
{
fast=fast.next;
if(fast!=null)
fast=fast.next;
else
return 0;
slow=slow.next;
if(slow==fast)
return 1;
}
return 0;
}
In C :
// Complete the has_cycle function below.
/*
* For your reference:
*
* SinglyLinkedListNode {
* int data;
* SinglyLinkedListNode* next;
* };
*
*/
bool has_cycle(SinglyLinkedListNode* head) {
struct SinglyLinkedListNode* slow=head, * fast=head;
while((slow!=NULL)&&(fast!=NULL)&&(fast->next!=NULL))
{
slow=slow->next;
fast=fast->next->next;
if(slow==fast)
return 1;
}
return 0;
}
In python3 :
"""
Check if linked list has cycle
head could be None as well for empty list
Node is defined as
class Node(object):
def __init__(self, data=None, next_node=None):
self.data = data
self.next = next_node
return 0 if no cycle else return 1
"""
def HasCycle(head):
slow = head
fast = head.next
count = 0
while slow and fast and count==0:
if slow == fast:
count=1
else:
slow= slow.next
fast = fast.next.next
if fast !=None or count == 1:
return(1)
else:
return(0)
View More Similar Problems
Reverse a linked list
Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio
View Solution →Compare two linked lists
You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis
View Solution →Merge two sorted linked lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C
View Solution →Get Node Value
This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t
View Solution →Delete duplicate-value nodes from a sorted linked list
This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -
View Solution →Cycle Detection
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer
View Solution →