How to realize Circular linked list by python
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[title]
Given a linked list, determine whether there are rings in the linked list.
If there is a node in the linked list that can be reached again by continuously tracking the next pointer, there is a ring in the linked list. To represent the rings in a given linked list, we use the integer pos to indicate where the tail of the linked list is connected to the linked list (the index starts at 0). If pos is-1, there are no rings in the linked list. Note: pos is not passed as a parameter, only to identify the actual situation of the linked list.
Returns true if there is a ring in the linked list. Otherwise, return false.
Advanced:
Can you solve this problem with O (1) (that is, constant) memory?
Example 1:
Input: head = [3Pere2Person0Lay 4], pos = 1 output: true explanation: there is a ring in the linked list, its tail is connected to the second node.
Example 2:
Input: head = [1jue 2], pos = 0 output: true explains: there is a ring in the linked list whose tail is connected to the first node.
Example 3:
Input: head = [1], pos =-1 output: false explanation: there are no rings in the linked list.
Tip:
The number of nodes in the linked list ranges from [0104] to 105 n ·m-n is the number of laps of the fast pointer relative to the slow pointer)
The starting position of the ring, the corresponding advance number is a, then: a = (m-2n) * c-b
Furthermore, we get: a = (m-2n + 1) * c + (c-b)
So, the slow pointer goes back to head, and both pointers move only one step at a time, so the meeting point is the starting point of the ring.
Code:
# Definition for singly-linked list.
# class ListNode:
# def _ _ init__ (self, x):
# self.val = x
# self.next = None
Class Solution:
Def detectCycle (self, head: ListNode)-> ListNode:
# find the meeting point
Fast, slow = head, head
Flag = False # No ring
While fast:
If fast and fast.next:
Fast = fast.next.next
Slow = slow.next
Else:
Break
If fast = = slow:
Flag = True
Break
If not flag:
Return None
# step = a + b + n * c (an is the number of elements before entering the ring; b is the number of remaining steps, c is the number of elements in the ring)
# 2 * step = a + b + m * c (m > ndirection m-n is the number of laps of the fast pointer relative to the slow pointer)
# = > a = (m-2n) * c-b
# a = (m-2n + 1) * c + (c-b)
# the slow pointer returns to head, and both pointers move only one step at a time, so the meeting point is the starting point of the ring
Slow = head
While fast! = slow:
Fast = fast.next
Slow = slow.next
Return fast
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