Linked List Cycle(E,Hash Table,Linked List,Two Pointers)

https://leetcode.com/problems/linked-list-cycle/description/

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?

 

1. Pyton

 

힐링 목적으로 찾은 Linked list, Easy 난이도.. 과아아연?!!

 

실패,, 처음 보는 유형의 문제다. 어려워 보이진 않은데 결국 답지정독..

 

기가 막히는 유형의 문제다. 이런게 재밌지

 

# 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:
        if not head or not head.next: return False

        slow = head
        fast = head.next

        while slow != fast :
            if not fast or not fast.next :
                return False
            slow = slow.next
            fast = fast.next.next
        return True

 

재밌는 친구다. 플로이드의 토끼/거북이 알고리즘이라고 한다. 역시 알고리즘이란.. 참..

 

2. C

 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
bool hasCycle(struct ListNode *head) {
    if(!head || !(head->next)) return false;
    
    struct ListNode *slow = head;
    struct ListNode *fast = head->next;
    
    while(slow != fast){
        if(!fast || !(fast->next)) return false;
        slow = slow->next;
        fast = fast->next->next;
    }
    return true;
}

나는 이렇게 파이썬과 똑같이 풀었는데,

다른 사람들 푼 답지 보다 보니까 웃긴 게 있었다 ㅋㅋㅋ

 

C
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
bool hasCycle(struct ListNode *head) {
    while(head) {
        if(head->val == '!@#') {
            return true;
        }
        head->val = '!@#';
        head = head->next;
    }
    return false;
}

이상한 문자를 추가하며 head 돌리다가 그놈 나오면 true 아니면 false return ㅋㅋㅋ

test case에서 없을만한 이상한 놈을 추가하고 돌리는 메커니즘인데 어찌보면 이것도 참 기발하다.

 

3. C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool hasCycle(ListNode *head) {
        if(!head ||!(head->next)) return false;
        ListNode* slow = head;
        ListNode* fast = head->next;

        while(slow!=fast){
            if(!fast||!(fast->next)) return false;
            fast = fast->next->next;
            slow = slow->next;
        }
        return true;

        
    }
};

 

비슷한 답변이다.