PHP中数字的递归和阶乘[重复]

Question

<?php  
function factorial_of_a($n)  
{  
    if($n ==0)  
    {  
        return 1;  
    }
    else   
    {    
        return $n * factorial_of_a( $n - 1 );  
    }  
}  
print_r( factorial_of_a(5) );
?>  

我的疑问是：

return $n * factorial_of_a( $n - 1 ) ;

在此语句中 - 当 $n = 5 和 $n - 1 = 4 时，它给出的结果为 20。 但是当我运行它时，答案 120 是怎么来的？嗯，120 是正确的答案......我不明白它是如何工作的。我改用for-loop，它工作正常。

Answer 1

factorial_of_a(5)

触发以下调用：

5 * factorial_of_a(5 - 1) ->
5 * 4 * factorial_of_a(4 - 1) ->
5 * 4 * 3 * factorial_of_a(3 - 1) ->
5 * 4 * 3 * 2 * factorial_of_a(2 - 1) ->
5 * 4 * 3 * 2 * 1 * factorial_of_a(1 - 1) ->
5 * 4 * 3 * 2 * 1 * 1

所以，答案是 120。

考虑阅读维基百科上的recursive function 文章。

另外，请阅读此相关主题：What is a RECURSIVE Function in PHP?

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但是答案 120 怎么来的？

好吧，这个函数会用$n - 1 调用自己，而$n - 1 不等于0。如果是，则函数实际上将结果返回给程序。所以它不会立即返回结果，而参数大于0。它被称为递归的“终止条件”。

Answer 2

它将以这种方式工作..

- factorial_of_a(5); 
// now read below dry run code from the bottom for proper understanding
// and then read again from top

   - if n = 0 ; false // since [n = 5]
   - else n*factorial_of_a(n-1); [return 5 * 24] 
   // here it will get 24 from the last line since 4*6 = 24 and pass 
   // it to the value of n i.e. **5** here will make it **120**

     - if n = 0 ; false [n = 4] 
     - else n*factorial_of_a(n-1); [return 4 * 6] 
     // here it will get 6 from the last line since 3*2 = 6 and pass it 
     // to the value of n i.e. **4** here will make it **24**

         - if n = 0 ; false [n = 3] 
         - else n*factorial_of_a(n-1); [return 3 * 2] 
         // here it will get 2 from the last line since 2*1 = 2 and pass it 
         // to the value of n i.e. **3** here will make it **6**

             - if n = 0 ; false [n = 2] 
             - else n*factorial_of_a(n-1); [return 2 * 1] 
             // here it will get 1 from the last line since 1*1 = 1 and pass it 
             // to the value of n i.e. **2** here

                 - if n = 0 ; false [n = 1] 
                 - else n*factorial_of_a(n-1); [return 1 * 1] 
                 // here it will get 1 from the last line and pass it 
                 // to the value of n i.e. **1** here

                     - if n = 0 ; true // since [n = 0] now it will return 1
                     - return 1;

Answer 3

你必须记住他是一个递归 当您调用 factorial_of_a(5) 时，它将执行为

factorial_of_a(5)  

// 5 * 24
5 * factorial_of_a(4)

  // 4 * 6
  4 * factorial_of_a(3)

     // 3 * 2
     3 * factorial_of_a(2)

         // 2 * 1
         2 * factorial_of_a(1)

             //will return 1 since it is your base condition
             1 * factorial_of_a(0)

Answer 4

要理解这一点，您需要清楚 递归 的概念。 递归意味着一次又一次地调用一个函数。 每个递归函数都有一个终止情况和一个递归情况。 终止 case 告诉函数何时停止，递归 case 再次调用函数本身。

在您的代码中，if 条件 $n == 0 标记了终止情况，即如果数字等于 0 并返回 1，则不进行任何计算。 else 部分是递归情况 [ $n*factorial_of_a($n-1) ]

Now i will explain how it works for $n = 5 :

Since $n is not equal to 0 then else statement is executed which gives 5 *factorial_of_a(4);

Now factorial_of_a(4) is called which gives 4 * factorial_of_a(3);

Now factorial_of_a(3) is called which gives 3 * factorial_of_a(2);

Now factorial_of_a(2) is called which gives 2 * factorial_of_a(1);

Now factorial_of_a(1) is called which gives 1 * factorial_of_a(0);

Now factorial_of_a(0) is called which gives 1

基本上是这样

factorial_of_a(5) = 5 * 4 * 3 * 2 * 1 = 120

结果就是这样！

希望对你有所帮助！