To print the sum of the numbers(not digits) in a string

Question

I was recently asked to write a function in java which will sum the numbers(not digits) in a string. eg if the string is abc35zz400tt15, the output should be 450.

this is what i wrote :

 public static void getSum(String a){
        String[] arr=a.split("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)");
        int sum=0;
        for(int i=0;i<arr.length;i++){
            if(Pattern.matches("[0-9]+", arr[i]))
        sum+=Integer.parseInt(arr[i]);
    }
    System.out.println(sum);
    }

Is there a more efficient way to do this as they didn't look satisfied with the above code.

Answer 1

I think it would be a bit simpler just to search for the pattern one at a time, rather than split the String.

Pattern pattern = Pattern.compile("[0-9]+"); //or you can use \\d if you want
Matcher matcher = pattern.matcher(a);

while(matcher.find()) {
    sum += Integer.parseInt(matcher.group());
}

Answer 2

If you are look forward efficiency regex matching and object allocations could be overkill for the task. You can scan the string backwards and accumulate the number:

int currentPower = 1;
int length = string.length();
int value = 0;

for (int i = string.length()-1; i >= 0; --i) {
  char curChar = string.charAt(i);
  if (curChar >= '0' && curChar <= '9') {
    value += (curChar - '0') * currentPower;
    currentPower *= 10;
  }
  else
    currentPower = 1;
}

Answer 3

Quick solution, without any usage of Matcher and Pattern:

    String line = "abc35zz400tt15";
    int sum = 0;

    String[] numbers = line.split("\\D");
    for (String digit : numbers) {
        if (digit.length() > 0) sum += Integer.valueOf(digit);
    }

Answer 4

Alternative solution: replace everything that is not a number with a + sign, and evaluate the result. That is not very efficient in Java, but there are other instances ( bash for example) where it would be the right thing to do. So I'm putting that suggestion here "for future visitors".

Clarification: I mean something like

echo abc35zz400tt15 | sed -E 's/[^0-9]+/\+/g' | sed 's/^\+//' | bc

That is

find one or more characters that are not a digit
replace all of them with a +
then strip the first `+` if it's the first character in the line
and evaluate the result

The above yields

450

as expected.

Note - bc doesn't like it when the string you feed it starts with a + - hence the need for the second sed .

Note 2 - the -E flag is used on Mac OSX to set "extended regular expressions". This gives special meaning ('one or more') to the + character - shorter than \\{1,}\\} ... I know the Mac one is subtly different than the gnu one, but don't know whether this is one of those occasions.

Answer 5

If you fancy Java 8, you can write it in a fairly expressive way:

int sum = Arrays.stream(input.split("\\D+"))
             .filter(s -> ! s.isEmpty())
             .mapToInt(Integer::parseInt)
             .sum();

Answer 6

Edit:: Whoops, read too fast, you wanted numbers, not digits. I believe someone has already answered above. I also have answered the question using my original (but incorrect) approach

Previous Answer

Look at Java's isDigit(char c) . It may look something like:

public static void getSum(string a) {
    int sum = 0;
    for(int i = 0; i < a.length(); i++) {
        if(Character.isDigit(a.charAt(i))) {
            sum += Integer.parseInt(a.charAt(i));
        }
    }
    System.out.println(sum);
}

New answer using an iterative approach

public static void getSum(String a) {
    int sum = 0;
    String num = "";
    for(int i = 0; i < a.length(); i++) {
        if(Character.isDigit(a.charAt(i))) {
            num = num + a.charAt(i);
        } else {
            if(!num.equals("")) {
                sum = sum + Integer.parseInt(num);
                num = "";
            }
        }
    }
    if(!num.equals("")) {
        sum = sum + Integer.parseInt(num);
    }
    System.out.println(sum);
}