Comp 170 Lab 2: lists of numbers

Comp 170 Lab 4 - Lists of Numbers (NumList), Sept 25

Goals

Introduction to Arrays
Introduction to for loops
Random numbers

Overview

This lab deals with a class NumList, which contains a list of numbers (ints), stored in an array called nums:
private int[] nums;Not all of the array is necessarily used; only the first currsize slots are (nums[0] through nums[currsize-1]). The variable currsize thus represents the length of the NumList, not nums.length. This way our NumList can easily grow in length, at least up until the hard limit of nums.length. The main task of this lab is to grow past that point.

I've given you a starter/demo project to get started. It starts out with nums.length = 5, and contains the following methods:

add(int y)
int get(int n)
void set(int n, int val)
int size()
void fillRandom(int max)
int sum()
void print()

The zipped project is here.

You are to do the following:

Implement NumList.max()

max() should return the largest element in the list. The catch is that potentially all of the numbers can be negative, so initializing your loop variable "max" to 0 may be too big. I started max for you in the NumList.java; there I suggest starting with max=nums[0]. Note that max is undefined for an empty NumList.

Make `add()` work even when `nums[]` is full

The trick here is to allocate a new array, say newnums, of larger size (doubling works, or increasing by 10, or increasing by 1.5 although that's trickier because of the floating-point conversions):
int newnums[] = new int[newsize];
Next you copy from nums[i] to newnums[i], for i<currsize. This is done with a straightforward for or while loop. Finally, you replace nums by newnums:
nums = newnums
does it. Note that there is no need to do anything with the "old" nums.

At this point, you now have space to add the new item, so you go ahead and do it. The overall add structure looks like:
    if (currsize == nums.length) {
        // do the space expansion as above to ensure room
}
// now add the usual way; one way or another we know there is space!
    nums[currsize] = y;
    currsize += 1;

Figure out how to add zero N times

The existing constructors set currsize = 0; that is, the NumList starts empty. But what if you want to start with N active slots? One way is to add zero a bunch of times;
for (int i = 0; i<N; i++) NL.add(0);
That works, and I did that in the Counter class, but it's kind of goofy; create a reasonable way to do this all at once. One approach is to add a method addMultiple(N), or grow(N). Another is to modify the second constructor that takes a parameter, so that
NumList NL = new NumList(10);
has the effect that the newly created NL now has currsize=10. Of course, it will also have a nums.length >= 10.

Use a NumList in a program

Here's the program, started in class Counter: generate 1000 random numbers in the range 0-9, and count how many times each number appears. Start with a NumList called counts, of currsize 10 and with all 10 positions initialized to 0. Then, for each new random number n,
n = rand.nextInt(10);
increment the value of slot n of counts; that is, of counts.get(n).