Scala中的广义懒排列

-1

我想创建一个通用的permutations函数。Scala中的广义懒排列

def permutation(size: Int): Stream[List[Int]] = ... 

permutation(1) # same as: for { x <- (1 to 10).toStream } yield List(x) 
permutation(2) # same as: for { x <- (1 to 10).toStream; 
            y <- (1 to 10).toStream; } yield List(x,y)

我能否动态嵌套在运行任意depth换推导行为。

来源

2015-09-26 zeronone

你有一个例子吗？ – Madoc

@Madoc我更新了问题。 – zeronone

我觉得这给你所要求的通用版本：

def permutation[X](iter: Iterable[X], len: Int) = { 
    def acc(items: List[X], count: Int): Stream[List[X]] = { 
    if (count == 0) Stream(items) else iter.toStream.flatMap(x => acc(x :: items, count - 1)) 
    } 

    acc(List.empty[X], len) 
}

使用范例：

scala> permutaion(1 to 3, 3) 
res4: Stream[List[Int]] = Stream(List(1, 1, 1), ?) 

scala> permutaion(1 to 3, 3).toList 
res5: List[List[Int]] = List(List(1, 1, 1), ?) 
res5: List[List[Int]] = List(List(1, 1, 1), List(2, 1, 1), List(3, 1, 1), List(1, 2, 1), List(2, 2, 1), List(3, 2, 1), List(1, 3, 1), List(2, 3, 1), List(3, 3, 1), List(1, 1, 2), List(2, 1, 2), List(3, 1, 2), List(1, 2, 2), List(2, 2, 2), List(3, 2, 2), List(1, 3, 2), List(2, 3, 2), List(3, 3, 2), List(1, 1, 3), List(2, 1, 3), List(3, 1, 3), List(1, 2, 3), List(2, 2, 3), List(3, 2, 3), List(1, 3, 3), List(2, 3, 3), List(3, 3, 3)) 

scala> permutation("abc", 3) 
res6: Stream[List[Char]] = Stream(List(a, a, a), ?) 

scala> permutation("abc", 3).toList 
res7: List[List[Char]] = List(List(a, a, a), List(b, a, a), List(c, a, a), List(a, b, a), List(b, b, a), List(c, b, a), List(a, c, a), List(b, c, a), List(c, c, a), List(a, a, b), List(b, a, b), List(c, a, b), List(a, b, b), List(b, b, b), List(c, b, b), List(a, c, b), List(b, c, b), List(c, c, b), List(a, a, c), List(b, a, c), List(c, a, c), List(a, b, c), List(b, b, c), List(c, b, c), List(a, c, c), List(b, c, c), List(c, c, c))

来源

2015-09-26 14:31:59 Shadowlands

哦，是的，这正是我想要的。解决方案非常优雅。 – zeronone

Scala中的广义懒排列

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