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:
[0, 104]
.-105 <= Node.val <= 105
pos
is -1
or a valid index in the linked-list.Source
def has_cycle(head):
'''Using set
Time Complexity: O(n)
Space Complexity O(n)
'''
if head is None:
return False
buff_set = {head}
current_node = head
while current_node.next:
current_node = current_node.next
if current_node in buff_set:
return True
buff_set.add(current_node)
return False
Can you solve it using O(1)
(i.e. constant) memory?
def has_cycle(head):
'''Using 2 pointers
space complexity can be reduced to O(1) by considering 2 pointers at different speed
- a slow pointer (moves 1 step at a time )
- a fast pointer (moves 2 steps at a time)
Cycle + case A: fast pointer is 1 step behind at time T
at T+1, they meet up
Cycle + Case B: fast pointer is 2 steps behind at time T
at T+1 they are in case A situation
at T+2 they meet up
Time Complexity: O(N+K) = O(n) with:
N = non-cyclic length
K = cyclic length
Space Complexity O(1)
'''
if head is None:
return False
slow = head
fast = head.next
while fast:
if fast.next is None:
return False
fast = fast.next.next
slow = slow.next
if slow == fast:
return True
return False