# 141. Linked List Cycle I

Given `head`

, the head of a linked list, determine if the linked list has a cycle in it.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next`

pointer. Internally, `pos`

is used to denote the index of the node that tail's `next`

pointer is connected to. **Note that ****pos**** is not passed as a parameter**.

Return `true`

* if there is a cycle in the linked list*. Otherwise, return `false`

.

**Example 1:**

**Input:** head = [3,2,0,-4], pos = 1

**Output:** true

**Explanation:** There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).

**Example 2:**

**Input:** head = [1,2], pos = 0

**Output:** true

**Explanation:** There is a cycle in the linked list, where the tail connects to the 0th node.

**Example 3:**

**Input:** head = [1], pos = -1

**Output:** false

**Explanation:** There is no cycle in the linked list.

**Constraints:**

- The number of the nodes in the list is in the range
`[0, 104]`

. `-105 <= Node.val <= 105`

`pos`

is`-1`

or a**valid index**in the linked-list.

Approach 1:

*Using hash set*

As we traverse the linked list , we basically store the address of each of the visited nodes to a hash set and then check if next pointer of current node points to any of the previously visited node, if yes then returns that node to which this current node points, else returns none

`def detectCycle(self, head):`

temp=head

hash_set=set()

index=0

while(temp!=None):

if(temp.next in hash_set):

return True

else:

hash_set.add(temp)

temp=temp.next

return False

Time :O(n)

Space :O(n)

Approach 2:

This is a solution which is better in terms of space and time , since we are already provided with the constraint that at most 1⁰⁴ nodes can be in the list, and in case of presence of a cycle , if we traverse the linked list until we find a null value , the iteration would go far beyond 1⁰⁴ , therefore this is a worthy condition to attribute to the presence of a cycle.

`def hasCycle(self, head):`

cnt=0

temp=head

while(temp!=None):

cnt+=1

if cnt>10**4:

return True

temp=temp.next

return False

Time :O(n)

Space:O(1)

Approach 3:

Floyd’s algo