By Behrouz A. Forouzan, Richard F. Gilberg

This moment version expands upon the forged, useful origin demonstrated within the first version of the textual content. a brand new four-part organizational constitution raises the flexibleness of the textual content, and all fabric is gifted in a simple demeanour observed by means of an array of examples and visible diagrams.

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**Additional resources for Data Structures: A Pseudocode Approach with C (2nd Edition)**

**Sample text**

Figure 1-7 shows the situation. void* p; int i; float f; p = &i; ... p = &f; p = &i p = &f p p i p f FIGURE 1-7 Pointers for Program 1-1 Program 1-1 uses a pointer to void that we can use to print either an integer or a floating-point number. Chapter 1 Basic Concepts 19 PROGRAM 1-1 Demonstrate Pointer to void 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 /* Demonstrate pointer to void. 500000 Program 1-1 Analysis The program is trivial, but it demonstrates the point. The pointer p is declared as a void pointer, but it can accept the address of an integer or floating-point number.

The big-O notation can be derived from f (n) using the following steps: 1. In each term, set the coefficient of the term to 1. Chapter 1 Basic Concepts 33 2. Keep the largest term in the function and discard the others. Terms are ranked from lowest to highest as shown below. logn n nlogn n2 n3 ... nk 2n n! For example, to calculate the big-O notation for (n + 1) 1 2 1 f ( n ) =n ----------------- = --- n + --- n 2 2 2 we first remove all coefficients. This gives us n2 + n which after removing the smaller factors gives us n2 which in big-O notation is stated as 2 O(f(n)) = O(n ) To consider another example, let’s look at the polynomial expression k f ( n ) = aj n + aj – 1 n k–1 2 + … + a2 n + a1 n + a0 We first eliminate all of the coefficients as shown below.

The general format is f (n) = efficiency The basic concepts are discussed in this section. 3. : Prentice Hall, 1988), xiii. Chapter 1 Basic Concepts 29 Linear Loops Let us start with a simple loop. We want to know how many times the body of the loop is repeated in the following code:4 for (i = 0; i < 1000; i++) application code Assuming i is an integer, the answer is 1000 times. The number of iterations is directly proportional to the loop factor, 1000. The higher the factor, the higher the number of loops.