打印字符串中字符的位置

Question

i have a method that prints the index of given character in a given String. If String doesn't contain given character the result is -1. My code looks like this:

public static int strpos(String text, char z) {

    int x = 0;

    char[] array = text.toCharArray();
    for (int i = 0; i < array.length; i++) {
        if (array[i] == z) {
            x = i;
            break;
        } else {
            x = -1;
        }
    }
    return x;
}

public static void main(String[] args) {
    System.out.println(strpos("abcdefghc", 'c'));
}

In this situation the result is 2 , and if i delete break , the result is 8 . How can i modyfi my code to get a result of both 2 and 8?

Answer 1

You can't get the result of both with a single return.
You can, however, return an ArrayList<Integer> .
You'll need to:

• Initialize the list at the head of the function.
• Each time you find a number, instead of breaking, add it your list.
• When you finish the search, return the list.

That's how you can return multiple values easily.

Also, why is the else there?

Answer 2

One solution would be to store the indices in a List<Integer> and change the method to return it:

public static List<Integer> strpos(String text, char z) {
    List<Integer> indices = new ArrayList<>();

    char[] array = text.toCharArray();

    for (int i = 0; i < array.length; i++) {
        if (array[i] != z) {
            continue;
        }

        list.add(i);
    }

    return indices.isEmpty() ? Arrays.asList(-1) : indices;
}

Answer 3

You can change the return type to List<Integer> and collect all the matching indexes in a list (or another collection of your choice). And in that case, instead of using -1 as a special value to indicate that a character doesn't exist in the string, return empty list instead, naturally.

Here's one way to implement that:

List<Integer> strpos(String text, char z) {
    return IntStream.range(0, text.length())
            .filter(i -> text.charAt(i) == z)
            .boxed()
            .collect(Collectors.toList());
}

Alternatively, without Java 8 streams, but benefitting from .indexOf :

List<Integer> strpos(String text, char z) {
    List<Integer> indexes = new ArrayList<>();
    int index = -1;
    while (true) {
        index = text.indexOf(z, index + 1);
        if (index == -1) {
            break;
        }
        indexes.add(index);
    }
    return indexes;
}