Java Fibonacci 系列 - 数组中的返回值
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Java Fibonacci Series - Return Values in Array
提问by corneria
I'm trying to recursively compute the fibonacci sequence to 100, store those returned values into an array using a the buildArray method, then print values stored in the array. I am getting a "cannot be resolved to a variable" compilation error when I try to print A[N] in the main method. I'm using longs because I'm computing the series up to 100, although I don't know if it's necessary to use longs.
我正在尝试将斐波那契数列递归计算为 100，使用 buildArray 方法将这些返回值存储到数组中，然后打印存储在数组中的值。当我尝试在 main 方法中打印 A[N] 时，出现“无法解析为变量”编译错误。我使用 longs 是因为我计算的序列最多为 100，虽然我不知道是否有必要使用 longs。
If I substitute F(N) for A[N] the code works, but I need to put the values into an array and print that array. Does this code even store the values in an array? I'm just starting java, thanks.
如果我用 F(N) 代替 A[N] 代码有效，但我需要将值放入数组并打印该数组。这段代码甚至将值存储在数组中吗？我刚刚开始java，谢谢。
public class MyFibonacci {
public static final int MAX = 100;
public static long[] buildArray(int MAX, int N) {
long[] A = new long[MAX];
A[0] = 0;
A[1] = 1;
for(N = 2; N < MAX; N++)
A[N] = F(N);
return A;
}
public static long F(int N) {
if(N == 0)
return 0;
if(N == 1)
return 1;
return F(N - 1) + F(N - 2);
}
public static void main(String[] args) {
for(int N = 0; N < MAX; N++)
System.out.println(N + " " + A[N]);
}
}
采纳答案by Bernhard Barker
The main problem is that you're never calling the buildArrayfunction.
主要问题是您从不调用该buildArray函数。
To get your code to work, you only need to add this to main:
要使您的代码正常工作，您只需要将其添加到main：
long[] A = buildArray(MAX, 0);
Some other things:
其他一些事情：
You can remove the parameter Nand just declare it in the function (or remove it all-together, see below).
您可以删除参数N并在函数中声明它（或一起删除它，见下文）。
You already have access to MAX, no need to pass it to the function.
您已经可以访问MAX，无需将其传递给函数。
The for-loop in buildArrayis rather inefficient, you can set up the array inside F.
in 的 for 循环buildArray效率很低，可以在里面设置数组F。
Given the below, Aas a class variable is cleaner than passing it around.
鉴于以下内容，A作为类变量比传递它更干净。
Finally, the code:
最后，代码：
static int MAX = 100;
static long[] A;
public static void buildArray()
{
A = new long[MAX+1];
F(MAX);
}
public static long F(int N)
{
long val;
if (N < 2)
val = N;
else if (A[N] != 0) // HEY! It's already calculated! Awesome! Just return it.
return A[N];
else
val = F(N-1) + F(N-2);
A[N] = val;
return val;
}
public static void main(String[] args)
{
buildArray();
for (int N = 0; N <= MAX; N++)
System.out.println(N + " " + A[N]);
}
回答by corneria
You have declared A[]within the scope of buildArray(int MAX, int N). As a result, A[]is not accessible outside of buildArray. You need to move your declaraction of long A[]to a class variable.
您已A[]在 范围内声明buildArray(int MAX, int N)。因此，A[]在buildArray. 您需要将您的声明移动long A[]到类变量。
Additionally, you actually need to run buildArrayfor the array to be constructed.
此外，您实际上需要buildArray为要构造的数组运行。
For future reference, I highly recommend using proper tabbing structures. It makes it much easier to see what's happening. I've edited your code (though it will have to be approved) to include this.
为了将来参考，我强烈建议使用适当的制表结构。它使查看正在发生的事情变得更加容易。我已经编辑了您的代码（尽管它必须获得批准）以包含此内容。
回答by d'alar'cop
Here's the code for what you need, I think:
这是您需要的代码，我认为：
public class MyFibonacci{
public static final int MAX = 100;
long[] A = new long[MAX];
public static long[] buildArray(int N){
A[0] = 0;
A[1] = 1;
for (N = 2; N < MAX; N++){
A[N] = F(N);
}
return A;
}
public static long F(int N)
{
if (N == 0) return 0;
if (N == 1) return 1;
return F(N-1) + F(N-2);
}
public static void main(String[] args)
{
buildArray(<some number - not sure where you get it from? N by the way in buildArray()>);
for (int N = 0; N < MAX; N++)
StdOut.println(N + " " + A[N]);
}
}
回答by Andrej
Since you can allocate array memory, it makes good sense to utilize it during calculation. Consider this method:
由于您可以分配数组内存，因此在计算过程中使用它很有意义。考虑这个方法：
public static long[] f_a(int n) {
long[] a = new long[n];
a[1] = 1;
for (int i = 2; i < n; i++)
a[i] = a[i-1] + a[i-2];
return a;
}
