SQL函数 - 阶乘

Question

我是SQL函数的初学者。 在SQL Server中为factorial创建函数的最佳方式是什么 - 说10！

Answer 1

这里是一个递归解决方案：

CREATE FUNCTION dbo.Factorial (@iNumber int) 
RETURNS INT 
AS 
BEGIN 
DECLARE @i int 

    IF @iNumber <= 1 
     SET @i = 1 
    ELSE 
     SET @i = @iNumber * dbo.Factorial(@iNumber - 1) 
RETURN (@i) 
END 

Answer 2

非递归的方式

;With Nums As 
(
select ROW_NUMBER() OVER (ORDER BY (SELECT 0)) AS RN 
FROM sys.objects 
) 
SELECT POWER(10.0, SUM(LOG10(RN))) 
FROM Nums 
WHERE RN <= 10 

和一个递归的方式

declare @target int 
set @target=10; 

WITH N AS 
    (SELECT 1 AS i, 
      1 AS f 

    UNION ALL 

    SELECT i+1, 
      f*(i+1) 
    FROM N 
    WHERE i < @target 
    ) 
SELECT f FROM N 
WHERE i=@target 

Answer 3

我知道我是一个有点晚了这里，但值得注意的是，Martin发布的递归方式不适用于0.

这将（原谅我，我有张贴代码问题）：

---

declare @target int=3; 

WITH N AS 
(SELECT 1 AS i, 
     1 AS f 
UNION ALL 
SELECT i+1, 
     f*(i+1) 
FROM N 
WHERE i < @target), 
N0 AS 
(SELECT f FROM N WHERE i=@target UNION SELECT 0) 
SELECT MAX(f) FROM N0 

而为方式，方法更快的版本：

declare @target int=5; 

WITH N AS 
(SELECT 1 AS i, 
     1 AS f 
UNION ALL 
SELECT i+1, 
     f*(i+1) 
FROM N 
WHERE i < @target), 
N0 AS 
(SELECT f FROM N WHERE i=@target UNION SELECT f=CASE WHEN @target=0 THEN 0 END) 
SELECT f FROM N0 
WHERE f>=0 

这是更快，因为我失去MAX（）函数，就像top 1一样，导致DISTINCT排序。

Answer 4

...我的Set-based的方法：

DECLARE @n int=11, @f bigint=1; 

WITH 
t(n,f) AS (SELECT TOP(@n) 
     ROW_NUMBER() OVER (ORDER BY (SELECT NULL))+1, 
     ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) * 
     (ROW_NUMBER() OVER (ORDER BY (SELECT NULL))+1) 
    FROM sys.all_columns 
    UNION SELECT 1, f=CASE WHEN @n=0 THEN 0 ELSE 1 END) 
SELECT @f=@f*f 
FROM t 
WHERE n%2=@n%2 OR f=0; 


SELECT @f AS FACTORIAL; 

Answer 5

- 迭代方法。 - 为什么迭代？它更简单快捷。 - 对于从0到20的@N，这给出了确切的结果。 - 21会发生溢出。

DECLARE @N Bigint = 20 
DECLARE @F Bigint = 1 
WHILE @N > 0 BEGIN 
    SET @F = @f*@n 
    SET @N = @N-1 
END 
SELECT @F AS FACTORIAL 

- 更改数据类型浮动，你可以得到阶乘高达170 - 171将导致溢出。 - 注意结果只能在有限的几个位置上准确。

DECLARE @N FLOAT = 170 
DECLARE @F FLOAT = 1 
WHILE @N > 0 BEGIN 
    SET @F = @f*@n 
    SET @N = @N-1 
END 
SELECT @F AS FACTORIAL 

- 本

Answer 6

试试这个

WITH MYCTE AS(
SELECT VAL=1,NUM =6 
UNION ALL 
SELECT VAL=VAL*NUM,NUM = (NUM -1) 
FROM MYCTE 
WHERE NUM > 1 
)     
SELECT VAL FROM MYCTE 

Answer 7

这里是一个其他的方法来calculate factorial value of an integer in SQL Server

create function sqlFactorial (@int int) 
returns int 
begin 
    declare @factorial bigint = 1 
    select @factorial = @factorial * i from dbo.NumbersTable(1,@int,1) 
    return @factorial 
end 

您需要使用SQL numbers table此解决方案。 Select语句将FROM部分中每行的已声明整数变量与有序整数值相乘

Answer 8

如果您对近似值没有问题，请使用Stirling's Approximation。

create table #temp (value int) 

insert into #temp values (5),(6),(7),(8) 

select 
    value, 
    sqrt(2*3.14*value)*power((value/2.718),value) --stirling's approx. 
from #temp 

请注意，如果需要，您将不得不提出0！的情况。