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 <= 105posis-1or a valid index in the linked-list.
Follow up: Can you solve it using O(1) (i.e. constant) memory?
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def intersection(self, head):
fast = head
slow = head
while fast and fast.next:
fast = fast.next.next
slow = slow.next
if fast == slow:
return slow
return None
def detectCycle(self, head) -> bool:
if not head:
return None
cycle = self.intersection(head)
if cycle is None:
return None
node1 = head
node2 = cycle
while node1 != node2:
node1 = node1.next
node2 = node2.next
return node1