Java Recursive Method它是如何工作的？

Question

I'm relatively new to Java programming and I've just started learning recursion, but I can't seem to figure out how this method works in my head.

   private static int mystery(int w) {
    {
        if (w < 0) return 0;
        int x = mystery (w-2);
        return w - x;
    }
}

Whenever a variable like 100 is put in, it outputs 50. When 200 is input, it outputs 100. When 2 is input, it outputs 2. When 25 is input, 13 is output. I'm not sure how this method works, and I'm trying to wrap my head around it.

The way I currently view it, if you put in 100, it'll bypass the first return statement since it is greater than 0. when it gets to the second line, it'll do 100-2, which brings in 98, then goes to the third line and does 100 - 98 = 2. Which is then returned to the original call.

I know I'm messing up on the second line of the method where the mystery (w-2) is. I assume it would bring back the result of w-2 to the beginning of the method again, and it would continue to do the method over and over again until w is smaller than 0, which should output 0 again regardless of the answer. But that's not what happens, and I don't know why.

Can anyone explain what is going on here?

Answer 1

What you are missing is that on the second line it doesn't just do w - 2, but calls itself with w - 2. It doesn't go further until the call returns. And the second call calls itself if w isn't < 0 and so on until you reach value lower than 0 and then return. The execution will go like this, if you visualize it:

mystery(10)
    > skip first line
    > x = mystery(8)
        > skip first line
        > x = mystery(6)
            > skip first line
            > x = mystery(4)
                > skip first line
                > x = mystery(2)
                    > skip first line
                    > x = mystery(0)
                        > skip first line
                        > x = mystery(-2)
                            > return 0
                        > return 0 - 0 (0)
                    > return 2 - 0 (2)
                > return 4 - 2 (2)
            > return 6 - 2 (4)
        > return 8 - 4 (4)
    > return 10 - 4 (6)

With example of w = 10. I hope you understand it better now.

Answer 2

   private static int mystery(int w) {
    {
        if (w < 0) return 0;
        int x = mystery (w-2);
        return w - x;
    }
}

Let's imagine that we call mystery(3) . What happens? w<0) is false, so we don't return 0. In the next line, we call some function called mystery using the value 3-2=1 as its argument. Despite the fact that this function we've called happens to be the same one we've just called, it's still an ordinary function call, and it returns a value. It does this by calling the function called mystery , this time using the value -1 as the argument. And this time w<0 is true, so we just return 0. Now we're back in the second call to mystery , and we've set x = 0. So that call returns w - 0 = 1. That puts us back in the first call, and now x = 1, so we return wx = 3-1 = 2.

You might want to take a few minutes and work through this using w=4 and see what you get - this will help you understand how the recursive calls work. After you've done this, I suggest you add a print statement or two in the function to tell you where you are and what's happening, and that'll also help - but do it on paper first.

Answer 3

The two given answers are excellent. Both focus on the way how to get a grasp of what recursion is. The problem with recursion is, that it is so unnatural to one who do not know what recursion is, or do not know someone who does. It's like a snake eating itself again and again.

The best way to understand recursion is to write down the calls to a recursive method, by noying the current state when it's called, and after the call write the result back. You stack up the calls and that's also the way to not used recursion at all.

So do not try too hard to understand recursion at first but first focus on the program flow. If you have seen enough recursions, it will come to you.