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Question

Formatted question description: https://leetcode.ca/all/141.html

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?

Algorithm

Two pointers, a slow pointer that walks one step at a time and a fast pointer that walks two steps at a time. If there is a ring in the linked list, the two pointers will definitely meet at the same position eventually.

Or, use a set for duplicate check.

Code

  • import java.util.HashSet;
import java.util.Set;

public class Linked_List_Cycle {

	/**
	 * Definition for singly-linked list.
	 * class ListNode {
	 *     int val;
	 *     ListNode next;
	 *     ListNode(int x) {
	 *         val = x;
	 *         next = null;
	 *     }
	 * }
	 */
    public class Solution {
        public boolean hasCycle(ListNode head) {
            if (head == null) {
                return false;
            }

            ListNode slow = head;
            ListNode fast = head;

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

            return false;
        }
    }

	public class Solution_extra_space {
	    public boolean hasCycle(ListNode head) {

	        Set<ListNode> hs = new HashSet<ListNode>();

	        ListNode current = head;

	        while (current != null) {

	            if (hs.contains(current)) {
	                return true;
	            }

	            hs.add(current);
	            current = current.next;
	        }

	        return false;

	    }
	}

}

############

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                return true;
            }
        }
        return false;
    }
}

  • // OJ: https://leetcode.com/problems/linked-list-cycle/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool hasCycle(ListNode *head) {
        ListNode *p = head, *q = head;
        while (q && q->next) {
            p = p->next;
            q = q->next->next;
            if (p == q) return true;
        }
        return false;
    }
};

  • # 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:
        slow = fast = head
        while fast and fast.next:
            slow, fast = slow.next, fast.next.next
            if slow == fast:
                return True
        return False

############

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

class Solution(object):
  def hasCycle(self, head):
    """
    :type head: ListNode
    :rtype: bool
    """
    fast = slow = head
    while fast and fast.next:
      fast = fast.next.next
      slow = slow.next
      if slow == fast:
        return True
    return False


  • /**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
 func hasCycle(head *ListNode) bool {
    slow, fast := head, head
    for fast != nil && fast.Next != nil {
        slow, fast = slow.Next, fast.Next.Next
        if slow == fast {
            return true
        }
    }
    return false
}

  • /**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function hasCycle(head: ListNode | null): boolean {
    const set = new Set<ListNode>();
    let node = head;
    while (node != null) {
        if (set.has(node)) {
            return true;
        }
        set.add(node);
        node = node.next;
    }
    return false;
}


  • /**
 * 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.next;
        if (slow == fast) {
            return true;
        }
    }
    return false;
};

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