I want to know if the way I format my for loops and if else statements is good style.

On the whole, yeah, looks good.

I recommend you use braces for one-liner blocks as well. In this instance, I'm not batting an eye at it—it's fine—but it's a good habit to learn.

I was wondering if I could do this withoutcount.

Yes, sincecount = j - i, you can elide it.

int sumNumbers(String str) {  int sum = 0;  for (int i = 0; i < str.length(); i++) {    if (Character.isDigit(str.charAt(i))) {      int j;      // start at i+1 because we know i has a digit      for (j = i + 1; j < str.length(); j++) {        if (!Character.isDigit(str.charAt(j))) {          break;        }      }      sum += Integer.parseInt(str.substring(i, j));      i = j;    }  }  return sum;}

Alternatively, if thebreak is a bit of an eye-sore, you could hoist the check into the for-header, but I'm not 100% it's an improvement:

int sumNumbers(String str) {  int sum = 0;  for (int i = 0; i < str.length(); i++) {    if (Character.isDigit(str.charAt(i))) {      int j;      // start at i+1 because we know i has a digit      for (j = i + 1; j < str.length() && Character.isDigit(str.charAt(j)); j++);      sum += Integer.parseInt(str.substring(i, j));      i = j;    }  }  return sum;}

Note:String.substring creates a new string since Java 7—in the reference implementation, at least, andString.toCharArray creates a new object too. If you want to avoid these allocations, you can do the char-to-digit conversion manually, but it's a bit messier, easy to muck up, and I feel that's out of scope for the exercise.

• \$\begingroup\$I like you're answer the best because it answered my individual questions even though I think Oscar Smith had a great answer as well.\$\endgroup\$

  dirtysocks45

  – dirtysocks45

  2017-10-29 09:29:21 +00:00

  CommentedOct 29, 2017 at 9:29
• \$\begingroup\$Also, the boolean condition in the second for loop is more of an eye sore to me. I find nothing wrong withbreak.\$\endgroup\$

  dirtysocks45

  – dirtysocks45

  2017-10-29 09:30:10 +00:00

  CommentedOct 29, 2017 at 9:30

There are a couple things you can do here to make things simpler. The first is to make your inner loop a while loop instead of a for loop. With this idea, you can build up each number char by char, usingCharacter.getNumericValue() which will return1 for'1' etc. With charVal, we can incrementally build up our number by multiplying the by 10 and adding the new digit. When we reach a non-digit, we add the current num to our sum, and set sum to 0. Since the code now only operates one char at a time, we don't needi, orj either. With these changes, your updated code is

public int sumNumbers(String str) {    int sum = 0;    int num = 0;    for (char ch : exampleString.toCharArray()) {        int digit = Character.getNumericValue(ch);        if (digit >= 0 && digit <= 9) {            num = num * 10 + digit;        } else {            sum += num;            num = 0;        }    }    return sum + num;}

• \$\begingroup\$How come you don’t initialize num to 1? And I think digit would be a better name.\$\endgroup\$

  dirtysocks45

  – dirtysocks45

  2017-10-28 18:49:09 +00:00

  CommentedOct 28, 2017 at 18:49
• 1
  \$\begingroup\$You initializenum to 0, because the first time you see a5 (eg), you wantnum*10 + 5 to equal 5, which means num needs to start at 0. It also means that when you don't hit number characters, you can addnum tosum without changing the answer.\$\endgroup\$

  Oscar Smith

  – Oscar Smith

  2017-10-28 18:52:14 +00:00

  CommentedOct 28, 2017 at 18:52
• \$\begingroup\$You also have some typos like missing;'s and and an extra) but when I started testing your program, I get different results. Some tests are now failing.sumNumbers("abc123xyz") returns0 instead of123.\$\endgroup\$

  dirtysocks45

  – dirtysocks45

  2017-10-28 19:00:00 +00:00

  CommentedOct 28, 2017 at 19:00
• \$\begingroup\$I edited your code to fix the above test case but cases where the integer is the last value in theString such assumNumbers("7 11") are returning the first integers like7 but forgetting about the last one like11.\$\endgroup\$

  dirtysocks45

  – dirtysocks45

  2017-10-28 19:16:26 +00:00

  CommentedOct 28, 2017 at 19:16
• 1
  \$\begingroup\$I just fixed thesumNumbers("7 11") case\$\endgroup\$

  Oscar Smith

  – Oscar Smith

  2017-10-28 19:21:45 +00:00

  CommentedOct 28, 2017 at 19:21

Your code looks fine and is short enough to be understood.

If you don't want to work on the level of individual characters, you can use regular expressions. Here is the code:

public static int sumNumbers(String str) {    Matcher m = Pattern.compile("[0-9]+").matcher(str);    int sum = 0;    while (m.find()) {        sum += Integer.parseInt(m.group());    }    return sum;}

This is not the fastest code, but I find it easy to read once you know that[0-9] meansa digit, and the+ means1 or more times.

• java
• strings
• integer
• modules