Linked List Cycle II LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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Linked List Cycle II Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

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 (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

Do not modify the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
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.

Linked List Cycle II Solutions

Time: O(n)
Space: O(1)

C++

class Solution {
 public:
  ListNode* detectCycle(ListNode* head) {
    ListNode* slow = head;
    ListNode* fast = head;

    while (fast && fast->next) {
      slow = slow->next;
      fast = fast->next->next;
      if (slow == fast) {
        slow = head;
        while (slow != fast) {
          slow = slow->next;
          fast = fast->next;
        }
        return slow;
      }
    }

    return nullptr;
  }
};

Java

public class Solution {
  public ListNode detectCycle(ListNode head) {
    ListNode slow = head;
    ListNode fast = head;

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

    return null;
  }
}

Python


class Solution:
  def detectCycle(self, head: ListNode) -> ListNode:
    slow = head
    fast = head

    while fast and fast.next:
      slow = slow.next
      fast = fast.next.next
      if slow == fast:
        slow = head
        while slow != fast:
          slow = slow.next
          fast = fast.next
        return slow

    return None

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