使用Java中的正则表达式拆分输入

Question

I am making a program that lets a user input a chemical for example C9H11N02. When they enter that I want to split it up into pieces so I can have it like C9, H11, N, 02. When I have it like this I want to make changes to it so I can make it C10H12N203 and then put it back together. This is what I have done so far. using the regular expression I have used I can extract the integer value, but how would I go about get C10, H11 etc..?

System.out.println("Enter Data");

Scanner k = new Scanner( System.in );
String input = k.nextLine();

String reg = "\\s\\s\\s";
String [] data;

data = input.split( reg );

int m = Integer.parseInt( data[0] );
int n = Integer.parseInt( data[1] );

Answer 1

It can be done using look arounds :

String[] parts = input.split("(?<=.)(?=[A-Z])");

Look arounds are zero-width, non-consuming assertions.

This regex splits the input where the two look arounds match:

• (?<=.) means "there is a preceding character" (ie not at the start of input)
• (?=[AZ]) means "the next character is a capital letter" (All elements start with AZ )

Here's a test, including a double-character symbol for some edge cases:

public static void main(String[] args) {
    String input = "C9KrBr2H11NO2";
    String[] parts = input.split("(?<=.)(?=[A-Z])");
    System.out.println(Arrays.toString(parts));
}

Output:

[C9, Kr, Br2, H11, N, O2]

---

If you then wanted to split up the individual components, use a nested call to split() :

public static void main(String[] args) {
    String input = "C9KrBr2H11NO2";
    for (String component : input.split("(?<=.)(?=[A-Z])")) {
        // split on non-digit/digit boundary
        String[] symbolAndNumber = component.split("(?<!\\d)(?=\\d)");
        String element = symbolAndNumber[0];
        // elements without numbers won't be split
        String count = symbolAndNumber.length == 1 ? "1" : symbolAndNumber[1];
        System.out.println(element + " x " + count);
    }
}

Output:

C x 9
Kr x 1
Br x 2
H x 11
N x 1
O x 2

Answer 2

Did you accidentally put zeroes into some of those formula where the letter "O" (oxygen) was supposed to be? If so:

"C10H12N2O3".split("(?<=[0-9A-Za-z])(?=[A-Z])");

[C10, H12, N2, O3]

"CH2BrCl".split("(?<=[0-9A-Za-z])(?=[A-Z])");

[C, H2, Br, Cl]

Answer 3

I believe the following code should allow you to extract the various elements and their associated count. Of course, brackets make things more complicated, but you didn't ask about them!

Pattern pattern = Pattern.compile("([A-Z][a-z]*)([0-9]*)");
Matcher matcher = pattern.matcher(input);
while (matcher.find()) {
    String element = matcher.group(1);
    int count = 1;
    if (matcher.groupCount > 1) {
        try {
            count = Integer.parseInt(matcher.group(2));
        } catch (NumberFormatException e) {
            // Regex means we should never get here!
        }
    }
    // Do stuff with this component
}