Q:

counting special characters with recursion

I'm trying to code this one up,but I don't get an expected result:
Given a string, compute recursively (no loops) the number of lowercase 'x' chars in the string. 
countX("xxhixx") → 4
countX("xhixhix") → 3
countX("hi") → 0
Here is my method:
public int countX(String str) {
    int count = 0;

    if(str.length() >= 1 ) {
        if(str.substring(0, 1).equals("x")) {
            str = str.substring(1, str.length());
            count = count + 1 + countX(str);
        }
    }
    else {
        str = str.substring(1, str.length());
        count = count + countX(str);
    }

    return count;
}

A:

You had the right idea, but I think you over complicated things. Just check explicitly if the first character is x (as you have), and only increment count in that case. Regardless of whether it was or wasn't, continue recursing on:
public static int countX(String str) {
    int count = 0;

    if (str.length() > 0) {
        if (str.substring(0, 1).equals("x")) {
            ++count;
        }

        str = str.substring(1, str.length());
        count += countX(str);

    }

    return count;
}

