141. Linked List Cycle

EasyLeetCode

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

How to solve the problem

Code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        #slow and fast point
        slow = head
        fast = head
        while fast and fast.next: # when fast pointer meet slow pointer
            slow = slow.next
            fast = fast.next.next
            if fast is slow:
                return True 
        return False

Complexity

• Time complexity: O(n)
• Space complexity: O(1)