# Print nodes of linked list at given indexes

Given a singly linked list which is sorted in ascending order and another singly linked list which is unsorted. The task is to print the elements of the second linked list according to the position pointed out by the data in the nodes of the first linked list. For example, if the first linked list is 1->2->5 then you have to print the 1st, 2nd and 5th elements of the second linked list.

Examples:

```Input: l1 = 1->2->5
l2 = 1->8->7->6->2->9->10
Output : 1->8->2
Elements in l1 are 1, 2 and 5.
Therefore, print 1st, 2nd and 5th elements of l2,
Which are 1, 8 and 2.

Input: l1 = 2->5
l2 = 7->5->3->2->8
Output: 5->8
```

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

Take two pointers to traverse the two linked lists using two nested loops. The outer loop points to the elements of the first list and the inner loop point to the elements of the second list respectively. In the first iteration of outer loop, the pointer to the head of the first linked list points to its root node. We traverse the second linked list until we reach the position pointed out by the node’s data in the first linked list. Once the required position is reached, print the data of the second list and repeat this process again until we reach the end of the first linked list.

Below is the implementation of the above approach:

 `// C++ program to print second linked list ` `// according to data in the first linked list ` `#include ` `using` `namespace` `std; ` ` `  `// Linked List Node ` `struct` `Node { ` `    ``int` `data; ` `    ``struct` `Node* next; ` `}; ` ` `  `/* Function to insert a node at the beginning */` `void` `push(``struct` `Node** head_ref, ``int` `new_data) ` `{ ` `    ``Node* new_node = ``new` `Node; ` ` `  `    ``new_node->data = new_data; ` `    ``new_node->next = (*head_ref); ` `    ``(*head_ref) = new_node; ` `} ` ` `  `// Fucntion to print the second list according ` `// to the positions referred by the 1st list ` `void` `printSecondList(Node* l1, Node* l2) ` `{ ` `    ``struct` `Node* temp = l1; ` `    ``struct` `Node* temp1 = l2; ` ` `  `    ``// While first linked list is not null. ` `    ``while` `(temp != NULL) { ` `        ``int` `i = 1; ` ` `  `        ``// While second linked list is not equal ` `        ``// to the data in the node of the ` `        ``// first linked list. ` `        ``while` `(i < temp->data) { ` `            ``// Keep incrementing second list ` `            ``temp1 = temp1->next; ` `            ``i++; ` `        ``} ` ` `  `        ``// Print the node at position in second list ` `        ``// pointed by current element of first list ` `        ``cout << temp1->data << ``" "``; ` ` `  `        ``// Increment first linked list ` `        ``temp = temp->next; ` ` `  `        ``// Set temp1 to the start of the ` `        ``// second linked list ` `        ``temp1 = l2; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``struct` `Node *l1 = NULL, *l2 = NULL; ` ` `  `    ``// Creating 1st list ` `    ``// 2 -> 5 ` `    ``push(&l1, 5); ` `    ``push(&l1, 2); ` ` `  `    ``// Creating 2nd list ` `    ``// 4 -> 5 -> 6 -> 7 -> 8 ` `    ``push(&l2, 8); ` `    ``push(&l2, 7); ` `    ``push(&l2, 6); ` `    ``push(&l2, 5); ` `    ``push(&l2, 4); ` ` `  `    ``printSecondList(l1, l2); ` ` `  `    ``return` `0; ` `} `

Output:

```5 8
```

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