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Rajni

Method 1 (Using Dummy Nodes) The strategy here uses a temporary dummy node as the start of the result list. The pointer Tail always points to the last node in the result list, so appending new nodes is easy. The dummy node gives tail something to point to initially when the result list is empty. This dummy node is efficient, since it is only temporary, and it is allocated in the stack. The loop proceeds, removing one node from either ‘a’ or ‘b’, and adding it to tail. When we are done, the result is in dummy.next. /*Program to alternatively split a linked list into two halves */ #include #include #include /* Link list node */ struct node { int data; struct node* next; }; /* pull off the front node of the source and put it in dest */ void MoveNode(struct node** destRef, struct node** sourceRef); /* Takes two lists sorted in increasing order, and splices their nodes together to make one big sorted list which is returned. */ struct node* SortedMerge(struct node* a, struct node* b) { /* a dummy first node to hang the result on */ struct node dummy; /* tail points to the last result node */ struct node* tail = &dummy; /* so tail->next is the place to add new nodes to the result. */ dummy.next = NULL; while(1) { if(a == NULL) { /* if either list runs out, use the other list */ tail->next = b; break; } else if (b == NULL) { tail->next = a; break; } if (a->data data) { MoveNode(&(tail->next), &a); } else { MoveNode(&(tail->next), &b); } tail = tail->next; } return(dummy.next); } /* UTILITY FUNCTIONS */ /*MoveNode() function takes the node from the front of the source, and move it to the front of the dest. It is an error to call this with the source list empty. Before calling MoveNode(): source == {1, 2, 3} dest == {1, 2, 3} Affter calling MoveNode(): source == {2, 3} dest == {1, 1, 2, 3} */ void MoveNode(struct node** destRef, struct node** sourceRef) { /* the front source node */ struct node* newNode = *sourceRef; assert(newNode != NULL); /* Advance the source pointer */ *sourceRef = newNode->next; /* Link the old dest off the new node */ newNode->next = *destRef; /* Move dest to point to the new node */ *destRef = newNode; } /* Function to insert a node at the beginging of the linked list */ void push(struct node** head_ref, int new_data) { /* allocate node */ struct node* new_node = (struct node*) malloc(sizeof(struct node)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print nodes in a given linked list */ void printList(struct node *node) { while(node!=NULL) { printf("%d ", node->data); node = node->next; } } /* Drier program to test above functions*/ int main() { /* Start with the empty list */ struct node* res = NULL; struct node* a = NULL; struct node* b = NULL; /* Let us create two sorted linked lists to test the functions Created lists shall be a: 5->10->15, b: 2->3->20 */ push(&a, 15); push(&a, 10); push(&a, 5); push(&b, 20); push(&b, 3); push(&b, 2); /* Remove duplicates from linked list */ res = SortedMerge(a, b); printf("\n Merged Linked List is: \n"); printList(res); getchar(); return 0; }

Help me choose my dress!!

Help me choose my dress!!