0141 - Linked List Cycle (Easy)

Problem Link

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

Problem Statement

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, 10^4].
• -10^5 <= Node.val <= 10^5
• pos is -1 or a valid index in the linked-list.

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

Approach 1: Two Pointers

C#

Written by @lonyehan

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public bool HasCycle(ListNode head) {
        ListNode slowPointer = head;
        ListNode quickPointer = head;
        
        if(head == null) return false;
        
        while(head != null) {
            // slow pointer, move one step each time.
            slowPointer = slowPointer.next;
            if(slowPointer == null) return false;
            
            // quick pointer, move two steps each time.
            quickPointer = quickPointer?.next?.next;
            if(quickPointer == null) return false;
            
            // slow pointer meets quick pointer means that there is a cycle in this linked list
            if(slowPointer == quickPointer) return true;
        }
        
        return false;
    }
}

Python

Written by @vale-c

# 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:
        slowPointer = head
        quickPointer = head
        
        if head == None:
            return False
        
        while head != None:
            # slow pointer, move one step each time.
            slowPointer = slowPointer.next
            if slowPointer == None:
                return False
            
            # quick pointer, move two steps each time.
            quickPointer = quickPointer.next.next if quickPointer.next != None else None
            if quickPointer == None:
                return False
            
            # slow pointer meets quick pointer means that there is a cycle in this linked list
            if slowPointer == quickPointer:
                return True
        
        return False

Java

Written by @vigneshshiv

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
// Time complexity: O(n), where n - # of nodes in the list
// Space complexity: O(1)
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null) return false;
        // Fast & Slow pointer
        ListNode slow = head;
        ListNode fast = head;
        // Fast Reference to check if it's not null, because it's traverse twice as fast as slow
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            // If both meet at the same node then there is a loop
            if (slow == fast) {
                return true;
            }
        }
        // If no loop, fast pointer at the end reached it's last node null pointer
        return false;
    }
}

JavaScript

Written by @radojicic23

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} head
 * @return {boolean}
 */
var hasCycle = function(head) {
    let slow = head
    let fast = head 

    while (fast && fast.next) {
        slow = slow.next;
        fast = fast.next;
        if (slow == fast) {
            return true;
        }
    }
    return false;
};

C++

Written by @radojicic23

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