#141

Linked List Cycle

pupil · 420 · lc easy +25 · failed · 54% accepted · 17,436 likes · top 46%

Description

Given head, the head of a linked list, determine whether the list contains a cycle.

A cycle exists if some node can be reached again by repeatedly following next pointers. The variable pos (not passed to your function) indicates which node the tail's next connects back to.

Return true if a cycle exists, otherwise 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.

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:

# 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: