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

Output:0

The easier method to find the cycle has been discussed in this post.

This method is based upon the idea of a circular race track. If there is a slow runner and a fast runner, eventually the fast runner will catch up the slow runner from behind. A similar analogy can be applied to a tortoise and a hare and hence the method is also called Tortoise and Hare method or the Floyd-Warshal algorithm.

The method is divided into 2 steps.

**STEP 1: Determine if there is a loop.**

The first phase/step of this method is to determine if there is actually any loop in the Linked List. If no loop is found, then we need to directly return null.

**STEP 2: Find the loop.**

- We initialize 2 pointers. Fast(hare) and slow(tortoise).
- Advance slow pointer by 1 step. Advance fast pointer by 2 steps.
- Keep performing the above step in a loop until both the pointers meet.
- As soon as they meet, start one more pointer from the beginning of the list.
- Move the slow pointer and the new pointer one step at a time.
- The place where they meet is the point where the loop starts in the linked list.

To have a better understanding of why this method works, have a look at the following image.

- Both hare and tortoise start from POINT 1.
- The hare covers distance
**‘F’**enters the loop and starts moving in the loop at POINT 2. - The hare runs for a while as he is fast and moves on path
**‘a’**. - The tortoise now enters the loop at POINT 2 after covering
**‘F’**and keeps on moving on path**‘a’**. While the hare is now on path**‘b’**. - The tortoise keeps on moving on path
**‘a’**and the hare eventually catches up with the tortoise at POINT 3. - Start a new pointer from POINT 1 and it runs at the same speed of tortoise.

We know that speed of hare is double the speed of tortoise.

Let us look at a little math now:-

2 * distance(tortoise) = distance(hare)

2 * (F + a) = F + a + b + a

2F + 2a = F + 2a + b

F = b

Hence POINT 2 is the point where loop starts.

The source-code for the above algorithm is given below.

// Template for a linked list node class ListNode { int value; ListNode next; } public class Solution { private ListNode getIntersect(ListNode head) { ListNode tortoise = head; ListNode hare = head; // A fast pointer will either loop around a cycle and meet the slow // pointer or reach the `null` at the end of a non-cyclic list. while (hare != null && hare.next != null) { tortoise = tortoise.next; hare = hare.next.next; if (tortoise == hare) { return tortoise; } } return null; } public ListNode detectCycle(ListNode head) { if (head == null) { return null; } // If there is a cycle, the fast/slow pointers will intersect at some // node. Otherwise, there is no cycle, so we cannot find an entrance to // a cycle. ListNode intersect = getIntersect(head); if (intersect == null) { return null; } // To find the entrance to the cycle, we have two pointers traverse at // the same speed -- one from the front of the list, and the other from // the point of intersection. ListNode ptr1 = head; ListNode ptr2 = intersect; while (ptr1 != ptr2) { ptr1 = ptr1.next; ptr2 = ptr2.next; } return ptr1; } }

*Time Complexity:* O(n)

*Space Complexity:* O(1)

A sample problem can be found here.

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