Given a linked list, return the node where the cycle begins. If there is no cycle, return
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. Note that
pos is not passed as a parameter.
Notice that you should not modify the linked list.
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.
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.
Input: head = , pos = -1 Output: no cycle Explanation: There is no cycle in the linked list.
- The number of the nodes in the list is in the range
-105 <= Node.val <= 105
-1or a valid index in the linked-list.
Follow up: Can you solve it using
O(1) (i.e. constant) memory?
We can have a hashmap to check whether if the node has been visited.
# 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: visited = set() while head != None: if head in visited: return head visited.add(head) head = head.next return None
- Time Complexity: O(N).
- Space Complexity: O(N).