# Find first node of loop in a linked list

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

Example :

```Input : Head of below linked list Output : Pointer to node 2```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

1. If a loop is found, initialize a 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 ` `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; ` `} `

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 means m+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
```

Method 3:
We can also use concept of hashing in order to detect first node of loop .The idea is simple just iterate over entire linked list and store node addresses in a set(C++ STL) one by one, while adding node address into the set check if it already contain that particular node address if not then add node address to set, if it is already present in the set then the current node is the first node of the loop.

 `// The below function take head of Linked List as ` `// input and return Address of first node in ` `// the loop if present else return NULL. ` ` `  `/* Definition for singly-linked list. ` ` ``* struct ListNode { ` ` ``*     int val; ` ` ``*     ListNode *next; ` ` ``*     ListNode(int x) : val(x), next(NULL) {} ` ` ``* };*/` `ListNode* detectCycle(ListNode* A) ` `{ ` ` `  `    ``// declaring map to store node address ` `    ``unordered_set uset; ` ` `  `    ``ListNode *ptr = A; ` ` `  `    ``// Default consider that no cycle is present ` `    ``while` `(ptr != NULL) { ` ` `  `        ``// checking if address is already present in map  ` `        ``if` `(uset.find(ptr) != uset.end())  ` `          ``return` `ptr; ` ` `  `        ``// if address not present then insert into the set ` `        ``else`  `            ``uset.insert(ptr);  ` `         `  `        ``ptr = ptr->next; ` `    ``} ` `    ``return` `NULL; ` `} ` `// This code is contributed by Pankaj_Joshi `

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