Linked List Cycle 2 (in Python)

Problem

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: 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.

 

Follow up:
Can you solve it without using extra space?

Solution

Approach 1 : Using A Hash Table

Approach 2 : Floyd's Cycle Detection Algorithm

I strongly recommend that you refer link below.

https://www.youtube.com/watch?v=LUm2ABqAs1w

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
 
class Solution(object):
    def Approach1(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        
        # Time complexity : O(N)
        # Space complexity : O(N)
        
        dic = {}
        
        while head :
            
            if head in dic :
                return head
            
            dic[head] = 1
            head = head.next
            
        return None
 
    def Approach2(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        
        # Time Complexity : O(N)
        # Space Complexity : O(1)
        
        slow = head
        fast = head
        
        while fast and fast.next :
            
            slow = slow.next
            fast = fast.next.next
            
            if slow == fast :
                
                slow = head
                
                while slow != fast :
                    slow = slow.next
                    fast = fast.next
                
                return slow
            
        return None

Source

https://leetcode.com/problems/linked-list-cycle-ii/