谐波序列递归 [英] Harmonic sequence recursion

问题描述

我确实掌握了递归的技巧(或者，我认为)，但是这个问题使我不寒而栗.我正在尝试返回1 + 1/2 + 1/3 + ... + 1/n，但是无论我尝试什么，该方法都将返回1.0.我一辈子都无法找出问题所在.

I'm really getting the hang of recursion (or so I think), but this problem is tripping me up. I'm trying to return 1 + 1/2 + 1/3 + ... + 1/n, but no matter what I try the method returns 1.0. I cannot for the life of me figure out what's wrong.

public static double harmonic(int n) {
    if(n == 1) {
        return 1;
    } else {
        return (1 / n) + (1 / harmonic(n - 1));
    }
}

推荐答案

好吧，对于一个，您不想返回(1/n)+(1/谐波(n-1))，但还需要使用 double 算术:

Well, for one, you don't want to return (1 / n) + (1 / harmonic(n - 1)), but also you need to use double arithmetic:

public static double harmonic(int n) {
    if(n == 1) {
        return 1.0;
    } else {
        return (1.0 / n) + harmonic(n - 1);
    }
}

如果将其保留为 1/谐波，则会完全返回另一个函数:

If you left it as 1 / harmonic you'd return another function entirely:

(1/n)+1/(1/(n-1)+1/(1/(n-2)+1/(...)))

(1 / n) + 1 / ( 1 / (n - 1) + 1 / ( 1 / (n - 2) + 1 / (...) ) )

这是一个非常令人困惑的功能，顺便说一句，但是我认为(我第三次编辑它)这次我做对了.

That is a very confusing function to figure out, btw, but I think (with my 3rd time editing it) I got it right this time.

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