Split a infix String to an array of String in java

Apparently each character is an apart token, except for consecutive digits with possibly a dot. So a simple solution would be iterating over the string, and then when you see a digit that is preceded by another digit (or the decimal separator, a dot), you add the character to the previous token, otherwise add it to a new token.

Here the code:

public static List<String> getTokens(String inputString) {
    List<String> tokens = new ArrayList<String>();
    // Add the first character to a new token. We make the assumption
    // that the string is not empty.
    tokens.add(Character.toString(inputString.charAt(0)));

    // Now iterate over the rest of the characters from the input string.
    for (int i = 1; i < inputString.length(); i++) {
        char ch = inputString.charAt(i); // Store the current character.
        char lch = inputString.charAt(i - 1); // Store the last character.

        // We're checking if the last character is either a digit or the
        // dot, AND if the current character is either a digit or a dot.
        if ((Character.isDigit(ch) || ch == '.') && (Character.isDigit(lch) || lch == '.')) {
            // If so, add the current character to the last token.
            int lastIndex = (tokens.size() - 1);
            tokens.set(lastIndex, tokens.get(lastIndex) + ch);
        }
        else {
            // Otherwise, add the current character to a new token.
            tokens.add(Character.toString(ch));
        }
    }
    return tokens;
}

Note that this method is faster than most regular expression methods.

You can use regex for parsing a mathamatical expression stored in a string.

expString.split("(?<=[-+*/\\(\\)])|(?=[-+*/\\(\\)])");

will do the trick for you.

Say,

String str = "100+(0.03*55)/45-(25+55)";
String[] outs = str.split("(?<=[-+*/\\(\\)])|(?=[-+*/\\(\\)])");
for (String element : outs)
{
    System.out.println(element);
}

Will give you an output of,

100
+
(
0.03
*
55
)
/
45
-
(
25
+
55
)

Please check my experiment @ http://rextester.com/QEMOYL38160

Here's an algorithm I would use:

start with an empty string array, and an empty string buffer

• walk from character 0 to character n
• for current character determine type (digit/period, open paren, close paren, math operators)
• if current character type is same as last character type
• add current character to buffer
• if not same, then put buffer into array of string, and start a new buffer

You need to use lookahead and look behind with split.

This works. Ofcourse, improve regex, if you want to include more elements.

public static void main(String[] args) {
    String input = "100+(0.03*55)/45-(25+55)";
    String test[] = input.split("((?<=[\\+\\-\\*\\/\\(\\)\\{\\}\\[\\]])|(?=[\\+\\-\\*\\/\\(\\)\\{\\}\\[\\]]))");
    System.out.println(Arrays.toString(test));
}

Update :

((?<=[a-z]]) , means it will split based on any character and include the character in the splited array as well after element.

(?=[a-z]) , means it will split based on any character and include the character in splited array before each element.

| , is the or operator between two regex.

[\\+\\-\\*\\/\\(\\)\\{\\}\\[\\]]) , is the regex to match to possible combinations

Please take a look at the answers to this other question.

This should do the trick:

Pattern p = Pattern.compile("(?:(\\d+)|([+-*/\\(\\)]))");
Matcher m = p.matcher("100+(0.03*55)/45-(25+55)");
List<String> tokens = new LinkedList<String>();
while(m.find())
{
  String token = m.group( 0 ); //group 0 is always the entire match   
  tokens.add(token);
}