递归方法 - Java
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Recursive method - Java
提问by MOTIVECODEX
Addition information:
补充资料：
Chip doesn't support multiplication, only addition. I should work around this problem by creating a recursive method, mult(), that performs multiplication of x and y by adding x to itself y times. Its arguments are x and y and its return value is the product of x and y. I should then write the method and a main() to call it.
芯片不支持乘法，只支持加法。我应该通过创建一个递归方法 mult() 来解决这个问题，该方法通过将 x 加到自身 y 次来执行 x 和 y 的乘法。它的参数是 x 和 y，它的返回值是 x 和 y 的乘积。然后我应该编写方法和一个 main() 来调用它。
It's pure logical thinking, but I get lost every time I try to think what to do.
这是纯粹的逻辑思维，但每次我试图思考该做什么时，我都会迷失方向。
I am stuck at the math part.. What I have, that doesn't work and I know the math is wrong, but I am not good at this:
我被困在数学部分..我所拥有的，那不起作用，我知道数学是错误的，但我不擅长这个：
public static void mult(int x, int y) {
x = 0;
y = 0;
if (y > 0) {
for (int i = 0; i < y; i++) {
x = x * (x * y);
return mult(x, y);
}
}
}
采纳答案by MOTIVECODEX
public static int mult(int x, int y) {
if (y == 0) {
return 0;
}
if (y > 0) {
return x + mult(x, y - 1);
} else {
return -x + mult(x, y + 1);
}
}
this was the solution by the way
顺便说一下，这是解决方案
回答by duffymo
When I hear "recursion", I expect to see two things:
当我听到“递归”时，我希望看到两件事：
- A function calling itself with modified argumentseach time.
- A stopping conditionright at the top that tells the function when to stop, avoiding an infinite stack.
- 每次使用修改后的参数调用自身的函数。
- 一个停止条件在顶部，它告诉函数权何时停止，避免了无限的堆栈。
So where are yours? Start with writing those down in words before you write code.
那么你的呢？在编写代码之前，先用文字写下来。
回答by subsub
... by adding x to itself y times.
...通过将 x 添加到自身 y 次。
You could actually do that, instead of multiplying. Oh, and maybe if you don't set both x and y to zero, you would have something to add ;-)
你实际上可以这样做，而不是乘法。哦，也许如果您不将 x 和 y 都设置为零，您将需要添加一些内容 ;-)
One last thing: If you want a recursive solution, you don't need the for-loop.
最后一件事：如果您想要递归解决方案，则不需要 for 循环。
回答by óscar López
All you need to remember is that a multiplication is a repeated addition (assumingthat both operands are >= 0), so we have:
你需要记住的是，乘法是一个重复的加法（假设两个操作数都是>= 0），所以我们有：
- The base case is when
yis zero - If
yis not zero, then addxone more time, and subtract1fromy
- 基本情况是什么时候
y为零 - 如果
y不为零，然后添加x一个更多的时间，并减去1从y
Notice that as long as yis positive, it'll eventually have a value of zero. So basically we keep adding xa total number of ytimes; this is what I mean:
请注意，只要y是正数，它最终的值就会为零。所以基本上我们一直在增加x总数y；这就是我的意思：
public static int mult(int x, int y) {
if (y == 0)
return 0;
return x + mult(x, y-1);
}
The same code can be written in a tail-recursivestyle, too - meaning: there's nothing to do after the recursive call returns, and this is important for certain languages that support a so-called tail-call optimization:
相同的代码也可以用尾递归风格编写- 意思是：递归调用返回后无事可做，这对于支持所谓的尾调用优化的某些语言很重要：
public static int mult(int x, int y, int accumulator) {
if (y == 0)
return accumulator;
return mult(x, y-1, x + accumulator);
}
The above will get called as follows, noticing that the last parameter is always initialized in zero:
上面将按如下方式调用，注意最后一个参数始终初始化为零：
mult(10, 5, 0)
=> 50
回答by Display Name
Java has no TCO by design, so using recursion for linear (not tree-like) processes is very bad idea. Especially for such task, which will most likely become a bottleneck in your program. Use loop instead.
Java 在设计上没有 TCO，因此对线性（不是树状）进程使用递归是非常糟糕的主意。特别是对于这样的任务，它很可能会成为您程序中的瓶颈。改用循环。
Oh, it must be recursive anyway? Looks like a homework task. Do it yourself then.
哦，无论如何它必须是递归的？看起来像是家庭作业。那就自己做吧。
回答by user2336315
One possibility is to use an accumulator which will store the current value of the multiplication. I replace missing statements by ??? :
一种可能性是使用一个累加器来存储乘法的当前值。我用 ??? 替换丢失的语句 ：
public static void main(String []args){
System.out.println(mult(2,5));
}
public static int mult(int x, int y) {
if(???) return ???;
else return multAcc(???,???,???);
}
private static int multAcc(int x, int y, int acc){
if(???) return ???;
else return multAcc(???, ???, ???);
}
