I need to add certain parts of the numerical string.
for example like.
036000291453
I want to add the numbers in the odd numbered position so like
0+6+0+2+1+5 and have that equal 14.
I tried the charAt(0)+charAt(2) etc, but it returns the digit at those characters instead of adding them. Thanks for your help.
Use charAt to get to get the char (ASCII) value, and then transform it into the corresponding int value with charAt(i) - '0'. '0' will become 0, '1' will become 1, etc.
Note that this will also transform characters that are not numbers without giving you any errors, thus Character.getNumericValue(charAt(i)) should be a safer alternative.
String s = "036000291453";
int total = 0;
for(int i=0; i<s.length(); i+=2) {
total = total + Character.getNumericValue(s.charAt(i));
}
System.out.println(total);
You can use Character.digit() method
public static void main(String[] args) {
String s = "036000291453";
int value = Character.digit(s.charAt(1), 10);
System.out.println(value);
}
Below code loops through any number that is a String and prints out the sum of the odd numbers at the end
String number = "036000291453";
int sum = 0;
for (int i = 0; i < number.length(); i += 2) {
sum += Character.getNumericValue(number.charAt(i));
}
System.out.println("The sum of odd integers in this number is: " + sum);
I tried the charAt(0)+charAt(2) etc, but it returns the digit at those
characters instead of adding them.
Character.getNumericValue(string.charAt(0));
Related
easy for usual numbers, but telephone numbers can start with 01-...., as an example, 01234
is basically 1234 for java, right?
So I can't divide by 10 recursively to find out how many digits there are.
is there any different way to find out how many digits there are ?
thanks in advance.
ps.: no regex if possible
Assume that the phone number is a string,
String pn = "049-4912394129" // that's a random value
then you could iterate in that string and check if a character is indeed a number
int count = 0;
for(char c : pn.toCharArray()){
if(Character.isDigit(c))
count++;
}
As you phone number is an int, you don't need to bother with regex and every locale phone number patterns.
Simply convert your intphone number to a String. Then you can easily get the string length.
int myIntNumber = 1234;
String myStringNumber = String.valueOf(myIntNumber);
int length = myStringNumber.length();
If you are talking about a number that represented by a String object then:
public int getNumberOfDigits(phoneNumber) {
int count = 0;
for (int i = 0, i < phoneNumber.length(); i++) {
if (Character.isDigit(phoneNumber.charAt(i))) {
count++;
}
}
System.out.println("Number of digits: " + count);
return count;
}
If you are talking about a number that represented by an int then simply
convert it to String before using it like so:
String phoneNumber = String.valueOf(phoneNumberInInt);
I assumed you DO want to count the zeros as a digit because you are not summarizing the value of them, so they do have a meaning when you talk about how many digits are there.
this could easily be done with a lambda
"049-4912394129".codePoints().filter( Character::isDigit ).count(); // 13
I am yet again stuck at the answer. This program prints the unique values but I am unable to get the sum of those unique values right. Any help is appreciated
public static void main(String args[]){
int sum = 0;
Integer[] numbers = {1,2,23,43,23,56,7,9,11,12,12,67,54,23,56,54,43,2,1,19};
Set<Integer> setUniqueNumbers = new LinkedHashSet<Integer>();
for (int x : numbers) {
setUniqueNumbers.add(x);
}
for (Integer x : setUniqueNumbers) {
System.out.println(x);
for (int i=0; i<=x; i++){
sum += i;
}
}
System.out.println(sum);
}
This is a great example for making use of the Java 8 language additions:
int sum = Arrays.stream(numbers).distinct().collect(Collectors.summingInt(Integer::intValue));
This line would replace everything in your code starting at the Set declaration until the last line before the System.out.println.
There's no need for this loop
for (int i=0; i<=x; i++){
sum += i;
}
Because you're adding i rather than the actual integers in the set. What's happening here is that you're adding all the numbers from 0 to x to sum. So for 23, you're not increasing sum by 23, instead, you're adding 1+2+3+4+5+....+23 to sum. All you need to do is add x, so the above loop can be omitted and replaced with a simple line of adding x to sum,
sum += x;
This kind of error always occures if one pokes around in low level loops etc.
Best is, to get rid of low level code and use Java 8 APIs:
Integer[] numbers = {1,2,23,43,23,56,7,9,11,12,12,67,54,23,56,54,43,2,1,19};
int sum = Arrays.stream(numbers)
.distinct()
.mapToInt(Integer::intValue)
.sum();
In this way there is barely any space for mistakes.
If you have an int array, the code is even shorter:
int[] intnumbers = {1,2,23,43,23,56,7,9,11,12,12,67,54,23,56,54,43,2,1,19};
int sumofints = Arrays.stream(intnumbers)
.distinct()
.sum();
So this is my first time commenting anywhere and I just really wanted to share my way of printing out only the unique values in an array without the need of any utilities.
//The following program seeks to process an array to remove all duplicate integers.
//The method prints the array before and after removing any duplicates
public class NoDups
{
//we use a void static void method as I wanted to print out the array without any duplicates. Doing it like this negates the need for any additional code after calling the method
static void printNoDups(int array[])
{ //Below prints out the array before any processing takes place
System.out.println("The array before any processing took place is: ");
System.out.print("{");
for (int i = 0; i < array.length; i++)
{
System.out.print(array[i]);
if (i != array.length - 1)
System.out.print(", ");
}
System.out.print("}");
System.out.println("");
//the if and if else statements below checks if the array contains more than 1 value as there can be no duplicates if this is the case
if (array.length==0)
System.out.println("That array has a length of 0.");
else if (array.length==1)
System.out.println("That array only has one value: " + array[0]);
else //This is where the fun begins
{
System.out.println("Processed Array is: ");
System.out.print( "{" + array[0]);//we print out the first value as it will always be printed (no duplicates has occured before it)
for (int i = 1; i < array.length; i++) //This parent for loop increments once the all the checks below are run
{
int check = 0;//this variable tracks the amount of times an value has appeared
for(int h = 0; h < i; h++) //This loop checks the current value for array[i] against all values before it
{
if (array[i] == array[h])
{
++check; //if any values match during this loop, the check value increments
}
}
if (check != 1) //only duplicates can result in a check value other than 1
{
System.out.print(", " + array[i]);
}
}
}
System.out.print("}"); //formatting
System.out.println("");
}
public static void main(String[] args)
{ //I really wanted to be able to request an input from the user but so that they could just copy and paste the whole array in as an input.
//I'm sure this can be done by splitting the input on "," or " " and then using a for loop to add them to the array but I dont want to spend too much time on this as there are still many tasks to get through!
//Will come back and revisit to add this if I remember.
int inpArray[] = {20,100,10,80,70,1,0,-1,2,10,15,300,7,6,2,18,19,21,9,0}; //This is just a test array
printNoDups(inpArray);
}
}
the bug is on the line
sum += i;
it should be
sum += x;
Alright, so my goal is to complete the following assignment:
"Design and implement an application that determines and prints the number of odd, even, and zero digits in an integer value read from the keyboard.
SPECIFICATION OF PROMPTS, LABELS AND OUTPUT : Your code should not have any prompt at all. The input to this program is a single integer . After the integer is read, the output consists of three lines. The first line consists of the number of odd digits in the integer followed by the label "odd digits". The second line consists of the number of even digits in the integer followed by the label "even digits". The third line consists of the number of zero digits in the integer followed by the label "zero digits". For example, if 173048 were read in, the output would be:
3 odd digits
3 even digits
1 zero digits
SPECIFICATION OF NAMES: Your application class should be called DigitAnalyst"
And the code I have produced is:
import java.util.Scanner;
public class DigitAnalyst{
public static void main(String[] args){
Scanner scan = new Scanner(System.in);
String num = scan.next();
int odd_count = 0;
int even_count = 0;
int zero_count = 0;
//input an int as a string, and set counter variables
int[] num_array = new int[num.length()];
//ready a array so we can so we can parse it sanely
for (int i =0; i < num.length(); i++)
{
num_array[i] = num.charAt(i);
}//fill the array with the values in the initial number using a loop
for ( int i=0;i< num_array.length; i++)
{
if (num_array[i] % 2 ==0)
{
if (num_array[i] ==0 )//the hell is going on here?
{
zero_count++;
}
else if (num_array[i] != 0)
{
even_count++;
}
}
else if (num_array[i] % 2 != 0)
{
odd_count++;
}
}//use this loop to check each part of the array
System.out.println(odd_count+ " odd digits");
System.out.println(even_count+" even digits");
System.out.println(zero_count+" zero digits");
}
}
And yet I keep getting the wrong output. More specifically, it returns the correct amount of odd numbers but it keeps counting 0 as an even and not as a zero.
I know where the problem is but I have no idea what is wrong, and I've spent a few hours on this.
If someone could point me in the right direction I'd be ectstatic.
When you encounter a problem that involves the manipulation of digits in an integer, the standard approach is to use an actual integer and the operator %, rather than strings. Instead of scan.next() use
int num = scan.nextInt();
And then you can do this:
do {
int digit = num % 10;
if ( digit == 0 ) {
zero_count ++;
} else if ( digit % 2 == 0 ) {
even_count ++;
} else {
odd_count ++;
}
num /= 10;
} while ( num > 0 );
The idea is that when you divide a number by 10, the remainder is exactly the rightmost digit, and the quotient will have all the other digits. That's simply how the decimal system works.
In this method you get the digit directly without calling any method, and you don't need any arrays.
If you assign the integer element with num.charAt(i) the ASCII value of the character is assigned and you get wrong results. In order to fix this, change
num_array[i] = num.charAt(i);
to
num_array[i] = Integer.parseInt(String.valueOf(num.charAt(i)));
or similar.
I'll give you some help here. First, charAt() returns the character at the index in the string, as a char data type. You're storing in an array of ints, which is assuming the numerical value of the character in the set, not the actual value.
Try this instead...
Change:
int[] num_array = new int[num.length()];
to:
char[] num_array = new char[num.length()];
and wrap all your num_array[i] references in your conditionals with:
Character.getNumericValue(num_array[i])
You should get your expected results.
Input = 12340
Output =
2 odd digits
2 even digits
1 zero digits
I am trying to take numbers from a users string (if it has numbers) and convert those numbers to their numerical value. I have the following code which takes in user input.
Ex: java Convert "55s" will just output the number 55, which i will store for later usage
{
char Element = 0;
double Sum = 0;
boolean Check = false;
for(String s: args) // taking in user input for command line
{
for (int i = 0; i<s.length(); i++)
{
Check = true;
Element = s.charAt(i); // converting the string into chars
Sum = convert_to_numb (Element, Check);
Check = false;
}
}
The input is a string in which i separate into chars and send it to my conversion functions. The idea i have follows
public static double convert_to_numb (char elem, boolean check) //trying to convert chars to numbers
{
char iter = elem;
double number = 0;
int count = 0;
while (check == true)
{
number = number + (iter - 48) * Math.pow(10,count);
System.out.println(iter);
count ++;
}
return number;
}
Here I am feeding in the chars to see if they're numbers and convert the actual numbers into their integer value. To try to clarify i would like to perform the following task given an example input of "55" covert it to 5*10^1 + 5*10^0 = 55. I would appreciate any help. Thanks.
Alright, I think I might know what you're trying to accomplish, though as others have mentioned it is a little unclear.
To address the code you just posted, I don't think it'll behave the way you expect. For starters, the Boolean variable 'Check' accomplishes nothing at the moment. convert_to_numb is only called while Check is true, so it's redundant.
Additionally, the sum isn't being stored anywhere as you loop through the string. Every time you obtain a sum, it overwrites the previous one.
Your convert_to_numb method is even more troubling; it contains an infinite loop. Since Check is always set to 'true', you essentially have a while(true) loop that will never end.
I'm going to assume that your objective here is to parse whichever Strings are input into the program looking for groups of consecutive digits. Then you want to store these groups of digits as integers, perhaps in an array if you find multiple.
Something like this might do the trick.
{
ArrayList<Integer> discovered = new ArrayList<Integer>();
for (String s : args) {
// contains previous consecutive digits (if any)
ArrayList<Integer> digits = new ArrayList<Integer>();
for (int i = 0; i < s.length(); i++) {
Character c = s.charAt(i);
// add digit to digit array
if (c.isDigit()) {
digits.add(c.getNumericValue())
}
// not a digit, so we clear the digit array
else {
// combine the array to form an integer
if (! digits.isEmpty()) {
int sum = 0;
int counter = 0;
for (Integer i : digits) {
sum += i * Math.pow(10, counter);
counter++;
}
discovered.add(sum);
digits.clear();
}
}
}
}
}
Note the use of ArrayLists and the Integer and Character wrapper classes. These all provide functionality that helps deal with edge-cases. For example, I'm not sure that (iter - 48) part would have worked in all cases.
Something like this:
public static void main(String[] args) {
String string = "55s";
String[] piece = string.split("[\\D]+");
for (int j = 0; j < piece.length; j++) {
if(piece[j].trim().length() > 0) {
System.out.println(piece[j]);
}
}
}
It will split your initial string, the rest you should do yourself.
I'm trying to convert a string filled with 16 digits into an array of ints where each index holds the digit of its respective index in the string. I'm writing a program where I need to do math on individual ints in the string, but all of the methods I've tried don't seem to work. I can't split by a character, either, because the user is inputting the number.
Here's what I have tried.
//Directly converting from char to int
//(returns different values like 49 instead of 1?)
//I also tried converting to an array of char, which worked,
//but then when I converted
//the array of char to an array of ints, it still gave me weird numbers.
for (int count = 0; count <=15; count++)
{
intArray[count] = UserInput.charAt(count);
}
//Converting the string to an int and then using division to grab each digit,
//but it throws the following error (perhaps it's too long?):
// "java.lang.NumberFormatException: For input string: "1234567890123456""
int varX = Integer.parseInt(UserInput);
int varY = 1;
for (count=0; count<=15; count++)
{
intArray[count]= (varX / varY * 10);
}
Any idea what I should do?
how about this:
for (int count = 0; count < userInput.length; ++count)
intArray[count] = userInput.charAt(count)-'0';
I think that the thing that is a bit confusing here is that ints and chars can be interpited as eachother. The int value for the character '1' is actually 49.
Here is a solution:
for (int i = 0; i < 16; i++) {
intArray[i] = Integer.valueOf(userInput.substring(i, i + 1));
}
The substring method returns a part of the string as another string, not a character, and this can be parsed to an int.
Some tips:
I changed <= 15 to < 16. This is the convetion and will tell you how many loop interations you will actually go throug (16)
I changed "count" to "i". Another convention...