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[Easy - wk6 - Q3] 141. Linked List Cycle

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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.

 

Example 1:

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

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.

 

Constraints:

  • The number of the nodes in the list is in the range [0, 104].
  • -105 <= Node.val <= 105
  • pos is -1 or a valid index in the linked-list.

 

Follow up: Can you solve it using O(1) (i.e. constant) memory?


관련자료

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namsoo님의 댓글

  • namsoo
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class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        nodes = set()
        node = head
        while node:
            if node in nodes:
                return True
            nodes.add(node)
            node = node.next
           
        return False

학부유학생님의 댓글

  • 학부유학생
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141. Linked List Cycle
Runtime: 90 ms, faster than 32.19% of Python3 online submissions for Linked List Cycle.
Memory Usage: 17.7 MB, less than 29.61% of Python3 online submissions for Linked List Cycle.

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

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        slow, fast = head, head
        
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
            if fast == slow: return True
        
        return False

Coffee님의 댓글

  • Coffee
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public class Solution {
    public boolean hasCycle(ListNode head) {
        ListNode dummy = new ListNode();
        dummy.next = head;
        
        Set<ListNode> isVisited = new HashSet<>();
        while(dummy.next != null){
            if(isVisited.contains(dummy.next)){
                return true;
            }
            isVisited.add(dummy.next);
            dummy = dummy.next;
        }
        
        return false;
        
    }
}
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