detect-linked-list-cycle.py

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

# Note: Do not modify the linked list.

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

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

class Solution(object):
    def detectCycle(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        slow=head
        fast=head
        while(slow and fast and fast.next):
            slow=slow.next
            fast=fast.next.next

            if(slow==fast):
                return 1
        return None

        
a=ListNode(20)
b=ListNode(4)
c=ListNode(5)
d=ListNode(15)
e=ListNode(20)

#a.next=b
#b.next=c
#c.next=d
#d.next=a

s=Solution()
print(s.detectCycle(a))