LeetCode-in-All

141. Linked List Cycle

Easy

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.

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

To solve this problem with the Solution class, we can use Floyd’s Tortoise and Hare algorithm, also known as the “slow and fast pointers” approach. Here are the steps:

1. Define the Solution class with a method hasCycle that takes the head of a linked list as input and returns a boolean indicating whether the linked list has a cycle.
2. Initialize two pointers, slow and fast, both pointing to the head of the linked list.
3. Iterate through the linked list using the slow and fast pointers.
4. Move the slow pointer one step at a time and the fast pointer two steps at a time.
5. If there is a cycle in the linked list, eventually the fast pointer will meet the slow pointer at some point.
6. If the fast pointer reaches the end of the list (None), it means there is no cycle, so return False.
7. If the fast and slow pointers meet, it means there is a cycle, so return True.

Here’s how the Solution class would look like in Python:

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        if not head or not head.next:
            return False
        
        slow = head
        fast = head.next
        
        while slow != fast:
            if not fast or not fast.next:
                return False
            slow = slow.next
            fast = fast.next.next
        
        return True

# Example usage:
solution = Solution()
head1 = ListNode(3)
head1.next = ListNode(2)
head1.next.next = ListNode(0)
head1.next.next.next = ListNode(-4)
head1.next.next.next.next = head1.next  # creates a cycle
print(solution.hasCycle(head1))  # Output: True

head2 = ListNode(1)
head2.next = ListNode(2)
head2.next.next = head2  # creates a cycle
print(solution.hasCycle(head2))  # Output: True

head3 = ListNode(1)
print(solution.hasCycle(head3))  # Output: False

This solution effectively uses Floyd’s Tortoise and Hare algorithm to detect whether there is a cycle in the linked list, satisfying the constraints provided in the problem statement.

Solution

# 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:
        if not head:
            return False
        fast = head.next
        slow = head
        while fast and fast.next:
            if fast == slow:
                return True
            fast = fast.next.next
            slow = slow.next
        return False

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