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# Problem:

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 = 1Output: trueExplanation: 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 = 0Output: trueExplanation: There is a cycle in the linked list, where the tail connects to the 0th node.`

Example 3:

`Input: head = , pos = -1Output: falseExplanation: 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.

Follow up: Can you solve it using `O(1)` (i.e. constant) memory?

# Solution:

`if not head or not head.next:    return False        slow, fast = head, head        while fast and fast.next:    slow = slow.next    fast = fast.next.next    if slow == fast:        return True            return False`

## Explanation —

• We use a fast and slow pointer approach here.
• The fast pointer travels twice as fast as the slow pointer.
• We return the node at which the two pointers meet.
• If the while loop terminates and there is no cycle, we return False.

## Time and Space Complexity —

Also, we require a constant amount of extra space as we have only used a couple of pointers.

Thus, if n is the length of the linked list,

Time Complexity: O(n)

Space Complexity: O(1)

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