##### Merge two sorted linked lists Hackerrank

You’re given the pointer to the head nodes of two sorted linked lists. The data in both lists will be sorted in ascending order. Change the next pointers to obtain a single, merged linked list which also has data in ascending order. Either head pointer given may be null meaning that the corresponding list is empty.

**Input Format **

You have to complete the Node* MergeLists(Node* headA, Node* headB) method which takes two arguments – the heads of the two sorted linked lists to merge. You should NOT read any input from stdin/console.

**Output Format **

Change the next pointer of individual nodes so that nodes from both lists are merged into a single list. Then return the head of this merged list. Do NOT print anything to stdout/console.

**Sample Input**

1 -> 3 -> 5 -> 6 -> NULL

2 -> 4 -> 7 -> NULL

15 -> NULL

12 -> NULL

NULL

1 -> 2 -> NULL

**Sample Output**

1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7

12 -> 15 -> NULL

1 -> 2 -> NULL

**Explanation **

1. We merge elements in both list in sorted order and output.

##### Solution

To merge two sorted linked lists, we can proceed by linearly traversing the lists and adding the node with the smaller value to the result and recursing for the remaining lists.

/* Merge two sorted lists A and B as one linked list Node is defined as struct Node { int data; struct Node *next; } */ Node* MergeLists(Node *headA, Node* headB) { if (headA == NULL && headB == NULL) return NULL; else if (headA == NULL) return headB; else if (headB == NULL) return headA;/* Insert Node at the end of a linked list head pointer input could be NULL as well for empty list Node is defined as class Node { int data; Node next; } */ Node MergeLists(Node list1, Node list2) { // This is a "method-only" submission. // You only need to complete this method Node dummy = new Node(); dummy.next=null; dummy.data=0; Node temp= dummy; while(true) { if(list1==null) { temp.next = list2; break; } else if(list2==null){ temp.next = list1; break; } else if(list1.data < list2.data) { temp.next=list1; list1=list1.next;""" Merge two linked lists head could be None as well for empty list Node is defined as class Node(object): def __init__(self, data=None, next_node=None): self.data = data self.next = next_node return back the head of the linked list in the below method. """ def MergeLists(xs, ys): if not ys: return xs elif not xs: return ys curr = result = None while xs and ys: if xs.data <= ys.data: if curr is None: curr = Node(xs.data) result = curr else: curr.next = Node(xs.data) curr = curr.next xs = xs.next else: if curr is None: curr = Node(ys.data) result = curr else: curr.next = Node(ys.data) curr = curr.next ys = ys.next if xs: curr.next = xs elif ys: curr.next = ys return result}

else{

temp.next=list2;

list2=list2.next;}

temp=temp.next;

}

return dummy.next;}

if(headA->data <= headB->data)

headA->next = MergeLists(headA->next, headB);else {

Node* temp = headB;

headB = headB->next;

temp->next = headA;

headA = temp;

headA->next = MergeLists(headA->next, headB);

}return headA;

// This is a "method-only" submission.

// You only need to complete this method

}