141. Linked List Cycle — LeetCode(Python)
I got you!
Problem:
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 <= 105posis-1or 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 FalseExplanation —
This is a very trivial problem.
- 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 —
Since, we are iterating over all elements in the array a constant number of times, we can say that our algorithm runs in linear time.
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)
