indexOf的递归实现

Question

I've already read many previous questions here and elsewhere, but I haven't found what I need. I need to write a recursive implementation of indexOf. The problem is that I can't use any local variables and have to give as input only a string and a char.

The method should return a value between 0 and the length of the string - 1 if the char has been found or -1 if it is not there. I know the actual 'indexOf' allows you to search for a string too, but this method is simplified.

I tried this but it's quite stupid since I used the real indexOf :

public static int indexOf(String s, char c){

    if(s.indexOf(c) < 0){       // I'd like to change this
        return -1;
    }

    if (s.length() == 0)        //base case #1
    {                           
        return -1;              
    } 
    else if (s.charAt(0) == c)  //base case #2
    {                           
        return 0;               
    }
    else {
        return 1 + indexOf(s.substring(1), c);
    }                                  
}

I saw this in particular, but is it possibile to write it without variables? Thanks

Answer 1

If you don't want local variables, you need to do the recursion in an internal method.

Advantage is that it's a lot faster, since it doesn't have to create new String objects, and the logic is tail-recursive, if used with a language that optimizes that.

public static int indexOf(String s, char c) {
    return indexOf0(s, c, 0);
}
private static int indexOf0(String s, char c, int index) {
    if (index == s.length())
        return -1;
    if (s.charAt(index) == c)
        return index;
    return indexOf0(s, c, index + 1);
}

Answer 2

The answer that you linked seems to be a good one... I recommend simply replacing the instances of the variable used in it with the method call the variable stores.

Below I simply edit the code:

public static int indexOf(char ch, String str) {
    // Returns the index of the of the character ch

    if (str == null || str.equals("")) {
        // base case: no more string to search; return -1
        return -1;
    } else if (ch == str.charAt(0)) {
        // base case: ch is at the beginning of str; return 0
        return 0; 
    }

    return indexOf(ch, str.substring(1)) == -1 ? -1 : 1 + indexOf(ch, str.substring(1));
}