141. Linked List Cycle - python3

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.

My Answer 1: Accepted (Runtime: 52 ms / Memory Usage: 17.2 MB)

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

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        if head is None or head.next is None:
            return False
        
        hA = head
        hB = head.next
        
        while hA!=hB:
            if hB is None or hB.next is None:
                return False
            hA = hA.next
            hB = hB.next.next
        
        return True

어제 배운 뒷목잡고 쓰러질뻔한 초간단 코드를 참고해서 시작했다

그 코드의 핵심은 한칸씩 움직이는 포인터와 먼저 앞서나가는 포인터라고 생각해서
hB는 걍 휙휙 지나가라고 .next.next를 해줌
head.next 가 없는 케이스에서 자꾸 오류 난리나서 조건문으로 제한을 많이 걸어뒀다.

Other Answer 1: (Runtime: 48 ms / Memory Usage: 17.6 MB)

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

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        s=set()
        while head is not None:
            if head not in s:
                s.add(head)
            else:
                return True
            head=head.next
        return False

해시set 이용한 방식
set 안에 있으면 무조건 true

막상 푼건 포인터지만 좀 때려맞춘거도 있어서...
set() 이용하는게 이해하기도 더 쉽고 간단한듯