正则表达式：奇数出现的奇数

Question

Found nothing specific to my problem in searches:

I have an Alphabet {a,b,c}, where I need to produce a set of strings that have an odd number of a's.

Valid : ababaccccc baaaccccc cab caabaaac

InValid : baac caacccb caabbbaac

Attempt:

\\b[bc]*a{3}[bc]*\\b but this is very limited.

Answer 1

以下正则表达式应该可行 。

\b[bc]*a(([bc]*a){2})*[bc]*\b

Answer 2

If you need solution without regex ie Java:

String arr[] = {"ababaccccc",  "baaaccccc" , "caabaaac", "baac", "caacccb", "caabbbaac"};   

for (String string : arr) {
            int counter = 0;
            for (int i = 0; i < string.length(); i++) {
                if (string.charAt(i) == 'a') {
                    counter++;
                }
            }
            if ((counter & 1) == 0) {
                System.out.println(string + " is invalid");
            } else {
                System.out.println(string + " is valid");
            }
        }

Answer 3

Wouldn't it be easier to

1. split the input string on whitespace
   
2. count the 'a's in every element
   
3. based on the outcome of count accept or reject ?