# Find first node of loop in a linked list

Write a function *findFirstLoopNode()* that checks whether a given Linked List contains loop. If loop is present then it returns point to first node of loop. Else it returns NULL.

Example :

Input : Head of bellow linked list Output : Pointer to node 2

We have discussed Floyd’s loop detection algorithm. Below are steps to find first node of loop.

1. If a loop is found, initialize slow pointer to head, let fast pointer be at its position.

2. Move both slow and fast pointers one node at a time.

3. The point at which they meet is the start of the loop.

`// C++ program to return first node of loop. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `struct` `Node { ` ` ` `int` `key; ` ` ` `struct` `Node* next; ` `}; ` ` ` `Node* newNode(` `int` `key) ` `{ ` ` ` `Node* temp = ` `new` `Node; ` ` ` `temp->key = key; ` ` ` `temp->next = NULL; ` ` ` `return` `temp; ` `} ` ` ` `// A utility function to print a linked list ` `void` `printList(Node* head) ` `{ ` ` ` `while` `(head != NULL) { ` ` ` `cout << head->key << ` `" "` `; ` ` ` `head = head->next; ` ` ` `} ` ` ` `cout << endl; ` `} ` ` ` `// Function to detect and remove loop ` `// in a linked list that may contain loop ` `Node* detectAndRemoveLoop(Node* head) ` `{ ` ` ` `// If list is empty or has only one node ` ` ` `// without loop ` ` ` `if` `(head == NULL || head->next == NULL) ` ` ` `return` `NULL; ` ` ` ` ` `Node *slow = head, *fast = head; ` ` ` ` ` `// Move slow and fast 1 and 2 steps ` ` ` `// ahead respectively. ` ` ` `slow = slow->next; ` ` ` `fast = fast->next->next; ` ` ` ` ` `// Search for loop using slow and ` ` ` `// fast pointers ` ` ` `while` `(fast && fast->next) { ` ` ` `if` `(slow == fast) ` ` ` `break` `; ` ` ` `slow = slow->next; ` ` ` `fast = fast->next->next; ` ` ` `} ` ` ` ` ` `// If loop does not exist ` ` ` `if` `(slow != fast) ` ` ` `return` `NULL; ` ` ` ` ` `// If loop exists. Start slow from ` ` ` `// head and fast from meeting point. ` ` ` `slow = head; ` ` ` `while` `(slow != fast) { ` ` ` `slow = slow->next; ` ` ` `fast = fast->next; ` ` ` `} ` ` ` ` ` `return` `slow; ` `} ` ` ` `/* Driver program to test above function*/` `int` `main() ` `{ ` ` ` `Node* head = newNode(50); ` ` ` `head->next = newNode(20); ` ` ` `head->next->next = newNode(15); ` ` ` `head->next->next->next = newNode(4); ` ` ` `head->next->next->next->next = newNode(10); ` ` ` ` ` `/* Create a loop for testing */` ` ` `head->next->next->next->next->next = head->next->next; ` ` ` ` ` `Node* res = detectAndRemoveLoop(head); ` ` ` `if` `(res == NULL) ` ` ` `cout << ` `"Loop does not exist"` `; ` ` ` `else` ` ` `cout << ` `"Loop starting node is "` `<< res->key; ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

Loop starting node is 15

**Output:**

Loop starting node is 15

**How does this approach work?**

Let slow and fast meet at some point after Floyd’s Cycle finding algorithm. Below diagram shows the situation when cycle is found.

We can conclude below from above diagram

Distance traveled by fast pointer = 2 * (Distance traveled by slow pointer) (m + n*x + k) = 2*(m + n*y + k) Note that before meeting the point shown above, fast was moving at twice speed. x --> Number of complete cyclic rounds made by fast pointer before they meet first time y --> Number of complete cyclic rounds made by slow pointer before they meet first time

From above equation, we can conclude below

m + k = (x-2y)*n Which meansm+k is a multiple of n.

So if we start moving both pointers again at **same speed** such that one pointer (say slow) begins from head node of linked list and other pointer (say fast) begins from meeting point. When slow pointer reaches beginning of loop (has made m steps), fast pointer would have made also moved m steps as they are now moving same pace. Since m+k is a multiple of n and fast starts from k, they would meet at the beginning. Can they meet before also? No because slow pointer enters the cycle first time after m steps.

** Method 2:**

In this method, a temporary node is created. The next pointer of each node that is traversed is made to point to this temporary node. This way we are using the next pointer of a node as a flag to indicate whether the node has been traversed or not. Every node is checked to see if the next is pointing to temporary node or not. In the case of the first node of the loop, the second time we traverse it this condition will be true, hence we return that node.

The code runs in O(n) time complexity and uses constant memory space.

Below is the implementation of the above approach:

// C++ program to return first node of loop

#include

using namespace std;

struct Node {

int key;

struct Node* next;

};

Node* newNode(int key)

{

Node* temp = new Node;

temp->key = key;

temp->next = NULL;

return temp;

}

// A utility function to print a linked list

void printList(Node* head)

{

while (head != NULL) {

cout << head->key << " ";
head = head->next;

}

cout << endl;
}
// Function to detect first node of loop
// in a linked list that may contain loop
Node* detectLoop(Node* head)
{
// Create a temporary node
Node* temp = new Node;
while (head != NULL) {
// This condition is for the case
// when there is no loop
if (head->next == NULL) {

return NULL;

}

// Check if next is already

// pointing to temp

if (head->next == temp) {

break;

}

// Store the pointer to the next node

// in order to get to it in the next step

Node* nex = head->next;

// Make next point to temp

head->next = temp;

// Get to the next node in the list

head = nex;

}

return head;

}

/* Driver program to test above function*/

int main()

{

Node* head = newNode(50);

head->next = newNode(20);

head->next->next = newNode(15);

head->next->next->next = newNode(4);

head->next->next->next->next = newNode(10);

/* Create a loop for testing */

head->next->next->next->next->next = head->next->next;

Node* res = detectLoop(head);

if (res == NULL)

cout << "Loop does not exist";
else
cout << "Loop starting node is " << res->key;

return 0;

}

**Output:**

Loop starting node is 15

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