使用Java中的递归的因子 [英] Factorial using Recursion in Java

问题描述

我正在使用Java：The Complete Reference这本书学习Java。
目前我正在研究Recursion主题。

I am learning Java using the book Java: The Complete Reference. Currently I am working on the topic Recursion.

请注意： stackoverflow上有类似的问题。我搜查了他们，但我找不到解决问题的方法。我对以下程序中的逻辑感到困惑。

Please Note: There are similar questions on stackoverflow. I searched them but I didn't find the solution to my question. I am confused with the logic in the following program.

如果我运行以下程序，它会产生正确的输出，但我不理解逻辑。

If I run the below program, it produces the correct output, but I didn't understand the logic.

• 我不理解以下行中的逻辑： result = fact（n-1）* n;


• 据我所知，如果我们传递n = 4的值，如下面的程序所示，


• 那么，3 * 4存储在结果ie，12。


• 再次，调用fact（n-1）。然后n变为3.


• 然后2 * 3存储在结果中，取代之前的12。



• 我想你明白了我感到困惑/困惑。

• I didn't understand the logic in the following line : result = fact(n-1) * n;
• From my knowledge, If we pass the value of n=4 as shown in the below program,
• Then, 3 * 4 is stored in the result i.e., 12.
• Again, fact(n-1) is called. Then n becomes 3.
• Then the 2 * 3 is stored in the result replacing the previous 12.

• I think you understood where I am stuck up/confused.

谢谢。

class Calculation
{
    int fact(int n)
    {
        int result;

       if(n==1)
         return 1;

       result = fact(n-1) * n;
       return result;
    }
}

public class Factorial
{
     public static void main(String args[])
     {
       Calculation obj_one = new Calculation();

       int a = obj_one.fact(4);
       System.out.println("The factorial of the number is : " + a);
     }
}

推荐答案

result 是 fact 方法的局部变量。因此，每次调用事实方法时，结果都存储在与前一个事实调用不同的变量中。

result is a local variable of the fact method. So each time the fact method is called, the result is stored in a different variable than the previous fact invocation.

因此，当使用3作为参数调用fact时，您可以想象它的结果是

So when fact is invoked with 3 as argument, you can imagine that its result is

 result3 = fact(2) * 3
 result3 = result2 * 3
 result3 = 1 * 2 * 3

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