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Two linked lists are having information of the same type in ascending order. Write down a module to merge them to a single linked list that is sorted
merge(struct node *p, struct node *q, struct **s)
{
struct node *z;
z = NULL;
if((x= =NULL) && (y = =NULL))
return;
while(x!=NULL && y!=NULL)
if(*s= =NULL)
*s=(struct link *)malloc(sizeof(struct node *z));
z=*s;
}
else
z-->link=(struct link *)malloc(sizeof(struct node *));
z=z-->link;
if(x-->data < y-->data)
z-->data=x-->data;
x=x-->link;
else if(x-->exp > y-->exp)
z-->data=y-->data;
y=y-->link;
else if(x-->data= =y-->data)
while(x!=NULL)
z link = struct link *malloc(sizeof(struct node *));
z=z link;
while(y!=NULL)
z-->link=NULL;
Queue is a linear data structure utilized in several applications of computer science. Such as people stand in a queue to get a specific service, several processes will wait in a q
Warnock's Algorithm An interesting approach to the hidden-surface problem was presented by Warnock. His method does not try to decide exactly what is happening in the scene but
Example of Back Face Detection Method To illustrate the method, we shall start with the tetrahedron (pyramid) PQRS of Figure with vertices P (1, 1, 2), Q (3, 2, 3), R (1,
(a) Describe the steps involved in the process of decision making under uncertainty. (b) Explain the following principles of decision making: (i) Laplace, (ii) Hurwicz. (c
Depth-first traversal A depth-first traversal of a tree visit a node and then recursively visits the subtrees of that node. Likewise, depth-first traversal of a graph visits
1. develop an algorithm which reads two decimal numbers x and y and determines and prints out wether x>y or y>x. the input values, x and y, are whole number > or equal to 0, which
difference between recursion and iteration
We have discussed already about three tree traversal methods in the earlier section on general tree. The similar three different ways to do the traversal -inorder , preorder, and p
A tree is a non-empty set one component of which is designated the root of the tree while the remaining components are partitioned into non-empty groups each of which is a subtree
Best Case: If the list is sorted already then A[i] T (n) = c1n + c2 (n -1) + c3(n -1) + c4 (n -1) = O (n), which indicates that the time complexity is linear. Worst Case:
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