Given a string find the first embedded occurrence of an integer

Question

This was asked in an interview:

Given in any string, get me the first occurence of an integer.

For example

Str98 then it should return 98

Str87uyuy232 -- it should return 87

I gave the answer as loop through the string and compared it with numeric characters, as in

if ((c >= '0') && (c <= '9'))

Then I got the index of the number, parsed it and returned it. Somehow he was not convinced. Can any one share the best possible solution?

Answer 1

With a regex, it's pretty simple:

String s = new String("Str87uyuy232");
Matcher matcher = Pattern.compile("\\d+").matcher(s);
matcher.find();
int i = Integer.valueOf(matcher.group());

_{(Thanks to Eric Mariacher)}

Answer 2

Using java.util.Scanner :

int res = new Scanner("Str87uyuy232").useDelimiter("\\D+").nextInt();

The purpose of a Scanner is to extract tokens from an input (here, a String). Tokens are sequences of characters separated by delimiters. By default, the delimiter of a Scanner is the whitespace, and the tokens are thus whitespace-delimited words.

Here, I use the delimiter \D+, which means "anything that is not a digit". The tokens that our Scanner can read in our string are "87" and "232". The nextInt() method will read the first one.

nextInt() throws java.util.NoSuchElementException if there is no token to read. Call the method hasNextInt() before calling nextInt(), to check that there is something to read.

Answer 3

There are two issues with this solution.

1. Consider the test cases - there are 2 characters '8' and '7', and they both form the integer 87 that you should be returning. (This is the main issue)

2. This is somewhat pedantic, but the integer value of the character '0' isn't necessarily less than the value of '1', '2', etc. It probably almost always is, but I imagine interviewers like to see this sort of care. A better solution would be

   if (Character.isDigit(c)) { ... }

There are plenty of different ways to do this. My first thought would be:

int i = 0;
while (i < string.length() && !Character.isDigit(string.charAt(i))) i++;
int j = i;
while (j < string.length() && Character.isDigit(string.charAt(j))) j++;
return Integer.parseInt(string.substring(i, j)); // might be an off-by-1 here

Of course, as mentioned in the comments, using the regex functionality in Java is likely the best way to do this. But of course many interviewers ask you to do things like this without libraries, etc...

Answer 4

String input = "Str87uyuy232";
Matcher m = Pattern.compile("[^0-9]*([0-9]+).*").matcher(input);
if (m.matches()) {
    System.out.println(m.group(1));
}