---
tags: leetcode
---

# [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)
---
# My Solution
## Solution 1
### The Key Idea for Solving This Coding Question
Use hash set to record the visited nodes. Once find a visited node in the hash set, there is a cycle in the linked list.
### C++ Code

/**
 * 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 == nullptr || head->next == nullptr) {
            return false;
        }

        unordered_set<ListNode *> visited;
        for (ListNode *it = head; it != nullptr; it = it->next) {
            auto iter = visited.find(it);
            if (iter != visited.end()) {
                // This node has been visited. Cycle detected!
                return true;
            }
            visited.insert(it);
        }

        return false;
    }
};

## Time Complexity
$O(n)$
$n$ is the number of nodes in the linked list referred by `head`.
## Space Complexity
$O(n)$
## Solution 2 (Floyd Cycle Detection Algorithm)
### The Key Idea for Solving This Coding Question
### C++ Code

/**
 * 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) {
        ListNode *fast = head, *slow = head;
        while (fast != nullptr && fast->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
            if (fast == slow) {
                return true;
            }
        }
        return false;
    }
};

## Time Complexity
$O(n)$
$n$ is the number of nodes in the linked list referred by `head`.
## Space Complexity
$O(1)$
## Solution 3
### The Key Idea for Solving This Coding Question
### C++ Code

/**
 * 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 == nullptr) {
            return false;
        }
        
        ListNode *curr = new ListNode(), *x = head->next;
        for (ListNode *it = head; it != nullptr; it = x) {
            if (it->next == curr) {
                return true;
            }
            x = it->next;
            it->next = curr;
        }
        
        return false;
    }
};

## Time Complexity
$O(n)$
$n$ is the number of nodes in the linked list referred by `head`.
## Space Complexity
$O(1)$
## Solution 4 (Brent Cycle Detection Algorithm)
### The Key Idea for Solving This Coding Question
### C++ Code

/**
 * 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 == nullptr) {
            return false;
        }
        
        ListNode *fast = head, *slow = head;
        unsigned long steps = 2;
        while (fast != nullptr) {
            for (unsigned long i = 0; i < steps && fast != nullptr; ++i) {
                fast = fast->next;
                if (fast == slow) {
                    return true;
                }
            }
            steps *= 2;
            slow = fast;
        }
        
        return false;
    }
};

## Time Complexity
$O(n)$
$n$ is the number of nodes in the linked list referred by `head`.
## Space Complexity
$O(1)$
# Reference
[探索 Floyd Cycle Detection Algorithm](https://hackmd.io/@eklipsorz/By-12SsEP)
# Miscellaneous
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