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Leetcode 142. Linked List Cycle II

Problem Statement

Given a linked list, return the node where the cycle begins. If there is no cycle, return nullptr.

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where the tail connects to form a cycle. If pos is -1, then there is no cycle.

Example:

• Input: head = [3,2,0,-4], pos = 1
• Output: node with value 2

Clarifying Questions

1. What should be returned if there’s no cycle in the linked list?
   • Return nullptr if there’s no cycle.
2. What is the range of values for the linked list nodes?
   • Each node’s value is an integer.
3. What about the size of the linked list?
   • The size of the linked list can range from 0 to 10^4.
4. Will there always be only one cycle in the linked list?
   • Yes, if a cycle exists, there’s only one cycle in the linked list.

Strategy

To detect the cycle and find the starting node of the cycle, we’ll use Floyd’s Tortoise and Hare algorithm:

1. Detect the Cycle:
   • Use two pointers, a slow pointer (tortoise) and a fast pointer (hare).
   • Initially, both pointers are at the head of the linked list.
   • Move the slow pointer one step at a time.
   • Move the fast pointer two steps at a time.
   • If the linked list has a cycle, the slow and fast pointers will eventually meet.
   • If there’s no cycle, the fast pointer will reach the end of the linked list (nullptr).
2. Find the Starting Node of the Cycle:
   • Once a cycle is detected, initialize another pointer at the head of the linked list.
   • Move both this new pointer and the slow pointer one step at a time.
   • The point at which they meet will be the start of the cycle.

Here’s the step-by-step code implementation:

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if (!head) return nullptr;
        
        ListNode *slow = head;
        ListNode *fast = head;
        
        // Detect if there is a cycle
        while (fast != nullptr && fast->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
            
            if (slow == fast) {
                // Cycle detected, find the entry point of the cycle
                ListNode *entry = head;
                while (entry != slow) {
                    entry = entry->next;
                    slow = slow->next;
                }
                return entry; // The start of the cycle
            }
        }
        
        return nullptr; // No cycle
    }
};

Time Complexity

• Time Complexity: O(n)
  • In the worst case, we traverse the entire linked list until the cycle is detected in O(n) time.
  • Finding the start of the cycle also requires O(n) time in the worst case.
  • Together, the operations are linear in terms of the number of nodes in the linked list.
• Space Complexity: O(1)
  • We use a constant amount of extra space regardless of the input size.

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