new question added

Author: arorarahul

README.md

@@ -135,6 +135,7 @@ TODO:
 - [Given a single linked list. Check if it has a loop](/linked-lists/question8.c)
 - [Find the starting node of cycle if cycle exists in a given linked list](/linked-lists/question9.c)
 - [Suppose there are two single linked lists both of which merge at the same point and become a single linked list. Find the node at which they merge](/linked-lists/question10.c)
+- [Alternating split of a given linked list](/linked-lists/question11.c)
 
 
 ## Some important concepts to solve algos better

linked-lists/question11.c

@@ -0,0 +1,65 @@
+/*
+Alternating split of a given linked list.
+
+We can use a counter a split at even and odd numbers and move them to new list, but then we have 
+dont have to create any new list but use the existing one and split it
+
+METHOD:
+take two pointers one pointing to head and one to head->next, then keep assigning them the next
+node as per the pattern until one of them reaches null. Make the other one also null in the end
+Time complexity: O(n)
+Space complexity: O(1)
+*/
+
+//METHOD
+#include <stdio.h>
+#include <stdlib.h>
+
+struct node{
+	int data;
+	struct node *link;
+};
+
+void makeList(struct node *t, int maxCounter, int mul){
+	int counter = 1;
+	while(counter <=maxCounter){
+		t->data = counter*mul;
+		if(counter == maxCounter){
+			t->link = NULL;	
+		}else{
+			t->link = (struct node *)malloc(sizeof(struct node));;	
+		}
+		t = t->link;
+		counter++;
+	}
+}
+
+void printList(struct node *head){
+	if(head){
+		printf("%d-->", head->data);
+		printList(head->link);
+	}
+	printf("\n");
+}
+
+void alternateList(struct node *head1, struct node *head2){
+	for(;head1 && head2 && head1->link && head2->link && head2->link->link;
+		head1=head1->link, head2=head2->link){
+		
+		head1->link = head2->link;
+		head2->link = head2->link->link;
+	}
+	head1->link = head2->link = NULL;
+}
+
+int main(){
+	struct node *head1 = (struct node *)malloc(sizeof(struct node));
+	struct node *t = head1;
+	makeList(t,8,10);
+	struct node *head2 = head1->link;
+	printList(head1);
+	alternateList(head1,head2);
+	printList(head1);
+	printList(head2);
+
+}