Thread: I/O List.java

Write a program that allows the user to enter a list of up to 100 integers. Please check for boundary conditions, for example, the user has to enter more than 0 integers but not more then 100. The end of the list is indicated by the special value -1. The program then reads more integers (max. 100), and, for every value, tells whether or not the value appears in the first list. The program ends when the user enters -1 (for the second time). For example, for the input

25 50 75 100 -1
99 25 30 -1

the program should respond:

99: not in list
25: yes, in list
30: not in list

So far I have the following

public class List
{
public static void main(String[]args)
{
// Input of the FirstList by the user.
int[] Firstlist = new int[100];
int i = 0;
int tempNum = StdIn.readInt();

while (tempNum != -1)
{
Firstlist[i] = tempNum;
tempNum = StdIn.readInt();
i = i + 1;
}
System.out.println(Firstlist[2]);

// Comparison
int [] Secondlist = new int[100];
int j = 0;
int tempNum2 = StdIn.readInt();

while (tempNum2 != -1)
{
Secondlist[j] = tempNum2;
tempNum2 = StdIn.readInt();
j = j + 1;
}
System.out.println(Secondlist[2]);

}
}

How do I compare the second list with the first list...Also note that Im using StdIn.readInt(); and I have all the files to use it...I got them from "introduction to programming an indeseplinary approach" website.

How do I compare the second list with the first list..

How do you want to compare the arrays?
If element by element, use a loop with an if test to compare the elements from each array.
Otherwise explain what you mean by comparing the arrays.

[QUOTE=Norm;60425]How do you want to compare the arrays?


The program has to tell you whether or not a value in the second list appears in the first list.

That sounds like a search. Take an element from the second list and search the first list.

When posting code, please wrap your code with[code=java]<YOUR CODE HERE>[/code] to get highlighting

Instead of using arrays, try using a LinkedList:

LinkedList<Integer> numbers1 = new LinkedList<Integer>();
LinkedList<Integer> numbers2 = new LinkedList<Integer>();
 
while(true){
System.out.println("Enter number");
int number = new Scanner(System.in).nextInt();
if(number == -1){
if(numbers1.size() > 0){
break;
}else{
System.out.println("Need atleast one element inside list");
}
}else{
if(numbers1.size() <= 100){
numbers1.add(number);
}else{
System.out.println("Can't have more then 100 elements, ending list");
break;
}
}
}
 
while(true){
System.out.println("Enter number");
int number = new Scanner(System.in).nextInt();
if(number == -1){
if(numbers2.size() > 0){
break;
}else{
System.out.println("Need atleast one element inside list");
}
}else{
if(numbers2.size() <= 100){
numbers2.add(number);
}else{
System.out.println("Can't have more then 100 elements, ending list");
break;
}
}
}
 
for(int i = 0; i < numbers2.size(); i++){
System.out.println(String.format("%d: %s", numbers2.get(i), numbers1.contains(numbers2.get(i)) ? "is in list" : "not in list"));
}

I might be a little off with the brackets because I wrote that inside here instead of Eclipse, but yeah, you get the point.