趣味编程：不使用循环对数组求和

Java代码

package hello;

public class TestA {
    public static int sum( int... numbers )
    {
       int total = 0;
       for ( int n : numbers )
          total += n;                 
       return total;
    }
    
    public static int sum2( int... numbers )
    {
       return sum2Helper(0, 0, numbers);
    }
    
    private static int sum2Helper( int total, int i, int... numbers)
    {
       return i == numbers.length ? total :
    	   sum2Helper(total + numbers[i], i + 1, numbers);
    }
    
    public static int sum3( int... numbers )
    {
       return sum3Helper(0, numbers);
    }
    
    private static int sum3Helper( int i, int... numbers)
    {
       return i == numbers.length ? 0 :
    	   numbers[i] + sum3Helper(i + 1, numbers);
    }
    
    public static void main( String args[] )
    {
    	System.out.println(sum(1, 2, 3, 4));
    	System.out.println(sum2(1, 2, 3, 4));
    	System.out.println(sum3(1, 2, 3, 4));
    }
}

//运行结果
//10
//10
//10

代码说明

1. sum函数使用普通的for each循环对数组求和。
   sum([1, 2, 3, 4]) = 0 + 1 + 2 + 3 + 4
2. sum2函数不使用循环而是用递归（左卷起）方式对数组求和。
   sum2([1, 2, 3, 4]) = (((0 + 1) + 2) + 3) + 4
3. sum3函数不使用循环而是用递归（右卷起）方式对数组求和。
   sum3([1, 2, 3, 4]) = 1 + (2 + (3 + (4 + 0)))

Haskell代码

sum2 :: Num a => [a] -> a
sum2 l = sum2Helper l 0
  where 
    sum2Helper [] a = a
    sum2Helper (x:xs) a = sum2Helper xs (a+x)


sum3 :: Num a => [a] -> a
sum3 l = sum3Helper l 0
  where
    sum3Helper [] a = a
    sum3Helper (x:xs) a = x + sum3Helper xs a

*Main> sum2 [1,2,3]
6
*Main> sum3 [1,2,3]
6

F#代码

let sum2 l =
    let rec sum2Helper l a =
        match l with
        | []    -> a
        | x::xs -> sum2Helper xs (a+x)
    sum2Helper l 0


let rec sum3 l =
    let rec sum3Helper l a =
        match l with
        | []    -> a
        | x::xs -> x + sum3Helper xs a
    sum3Helper l 0

> sum2 [1;2;3];;
val it : int = 6
> sum3 [1;2;3];;
val it : int = 6