LeetCode 141. Linked List Cycle

141. Linked List Cycle

 

Given head, the head of a linked list, determine if the linked list has a cycle in it.

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.

Return true if there is a cycle in the linked list. Otherwise, return false.

 

주어진 연결리스트에 사이클이 있다면 true를 없다면 false를 반환하라

Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: true
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: true
Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.


Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

 

Solution1 : 

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        if head==None:
            return False
        if head.next==None:
            return False
        slow=head
        fast=head
        while fast!=None and fast.next!=None:
            fast=fast.next.next
            slow=slow.next
            if slow==fast:
                return True
        return False

 

Solution2 :

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        """
        O(n) time O(1) space
        slow = head
        fast = head
        while(fast != None and fast.next != None):
            slow = slow.next
            fast = fast.next.next
            
            if slow == fast:
                return True
            
        return False
        """
        
        # O(n) time and O(n) space
        dictionary = collections.defaultdict(ListNode)
        while(head):
            if head in dictionary:
                return True
            dictionary[head] = True
            head = head.next
        return False