142. Linked List Cycle II

• python3

Problem

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

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 (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

Do not modify the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
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?

Pseudocode

1. Detect the Cycle.
2. If there is a cycle, reset one pointer to head, and increment each pointers by one.
3. When they meet, that point will be the start node of the cycle.

Floyd의 토끼와 거북이 알고리즘 (Floyd's Tortoise and Hare Algorithm)

참고
https://blog.naver.com/occidere/222260962156

하여 추후 정리...

Code

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

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:

        if not head:
            return None

        slowNode = head
        fastNode = head

        while fastNode and fastNode.next:
            slowNode = slowNode.next
            fastNode = fastNode.next.next
            if slowNode == fastNode:
                break
        else:
            return None

        slowNode = head
        while slowNode != fastNode:
            fastNode = fastNode.next
            slowNode = slowNode.next
        

        return slowNode

Result

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