我編寫了一個枚舉給定列表的所有排列的函數。您如何看待下面的代碼？枚舉Scala中排列的代碼
def interleave(x:Int, l:List[Int]):List[List[Int]] = {
l match {
case Nil => List(List(x))
case (head::tail) =>
(x :: head :: tail) :: interleave(x, tail).map(head :: _)
}
}
def permutations(l:List[Int]):List[List[Int]] = {
l match {
case Nil => List(List())
case (head::tail) =>
for(p0 <- permutations(tail); p1 <- interleave(head, p0)) yield p1
}
}
給定一個Seq，可以通過調用permutations方法已經有排列。
scala> List(1,2,3).permutations.mkString("\n")
res3: String =
List(1, 2, 3)
List(1, 3, 2)
List(2, 1, 3)
List(2, 3, 1)
List(3, 1, 2)
List(3, 2, 1)
此外還有用於combinations的方法：
scala> List(1,2,3).combinations(2).mkString("\n")
res4: String =
List(1, 2)
List(1, 3)
List(2, 3)
關於你實現我要說三兩件事：
（1）它好看
（2）提供一個迭代器（這是std collections方法，允許丟棄元素）。否則，你可以得到1000的列表！可能不適合記憶的元素。
scala> val longList = List((1 to 1000):_*)
longList: List[Int] = List(1, 2, 3,...
scala> permutations(longList)
java.lang.OutOfMemoryError: Java heap space
at scala.collection.immutable.List.$colon$colon(List.scala:67)
at .interleave(<console>:11)
at .interleave(<console>:11)
at .interleave(<console>:11)
（3）您應刪除重複的排列（由Luigi觀察到），因爲：
scala> permutations(List(1,1,3))
res4: List[List[Int]] = List(List(1, 1, 3), List(1, 1, 3), List(1, 3, 1), List(1, 3, 1), List(3, 1, 1), List(3, 1, 1))
scala> List(1,1,3).permutations.toList
res5: List[List[Int]] = List(List(1, 1, 3), List(1, 3, 1), List(3, 1, 1))
我認爲這樣的功能已經存在於標準庫中：Seq.permutations。那麼爲什麼要重新發明輪子？
考慮這裏的區別是：您的版本
scala> permutations(List(1,1,2)) foreach println
List(1, 1, 2)
List(1, 1, 2)
List(1, 2, 1)
List(1, 2, 1)
List(2, 1, 1)
List(2, 1, 1)
的參考版本：
scala> List(1,1,2).permutations foreach println
List(1, 1, 2)
List(1, 2, 1)
List(2, 1, 1)
哇。如果在不閱讀/理解文檔的情況下使用，參考實現可能實際上會破壞算法通常，你會期望'x.permutations.size == faculty（x.size）'。 – Raphael
I猜你正在練習你的Scala編程技巧。這裏是另一個，其中的想法是採取不同的元素作爲序列的頭，並通過filter刪除重複。由於O（n）+ O（n或可能n^2）+ O（n）* P（n-1）由O（n）* P（n-1）支配，所以代碼的複雜性應該是很好的。其中P（n）是置換數且不能改進。
def permute(xs:List[Int]):List[List[Int]] = xs match {
case Nil => List(List())
case head::tail => {
val len = xs.length
val tps = (0 to len-1).map(xs.splitAt(_)).toList.filter(tp => !tp._1.contains(tp._2.head))
tps.map(tp => permute(tp._1:::tp._2.tail).map(tp._2.head :: _)).flatten
}
}
也許這線程已經非常飽和的，但我想我會扔我的解決辦法混進去：
假設沒有重複元素：
def permList(l: List[Int]): List[List[Int]] = l match {
case List(ele) => List(List(ele))
case list =>
for {
i <- List.range(0, list.length)
p <- permList(list.slice(0, i) ++ list.slice(i + 1, list.length))
} yield list(i) :: p
}
隨着重複元件，防止重複（不如漂亮）：
def permList(l: List[Int]): List[List[Int]] = l match {
case List(ele) => List(List(ele))
case list =>
for {
i <- List.range(0, list.length)
val traversedList = list.slice(0, i)
val nextEle = list(i)
if !(traversedList contains nextEle)
p <- permList(traversedList ++ list.slice(i + 1, list.length))
} yield list(i) :: p
}
它可能不是最「list-y」，給定它在列表中使用了切片和索引，但它非常簡潔並且稍微有點不同。它通過挑選列表中的每個元素並計算剩餘內容的排列，然後將單個元素連接到每個排列上。如果有更習慣的方式來做到這一點，我很樂意聽到它。
我認爲，我的解決方法是比別人
def withReplacements(chars: String, n: Int) {
def internal(path: String, acc: List[String]): List[String] = {
if (path.length == n) path :: acc else
chars.toList.flatMap {c => internal(path + c, acc)}
}
val res = internal("", Nil)
println("there are " + res.length + " " + n + "-permutations with replacement for " + chars + " = " + res)
} //> withReplacements: (chars: String, n: Int)Unit
def noReplacements(chars: String, n: Int) {
//val set = chars.groupBy(c => c).map {case (c, list) => (c -> list.length)}.toList
import scala.collection.immutable.Queue
type Set = Queue[Char]
val set = Queue[Char](chars.toList: _*)
type Result = Queue[String]
// The idea is that recursions will scan the set with one element excluded.
// Queue was chosen to implement the set to enable excluded element to bubble through it.
def internal(set: Set, path: String, acc: Result): Result = {
if (path.length == n) acc.enqueue(path)
else
set.foldLeft(acc, set.dequeue){case ((acc, (consumed_el, q)), e) =>
(internal(q, consumed_el + path, acc), q.enqueue(consumed_el).dequeue)
}. _1
}
val res = internal(set, "", Queue.empty)
println("there are " + res.length + " " + n + "-permutations without replacement for " + set + " = " + res)
} //> noReplacements: (chars: String, n: Int)Unit
withReplacements("abc", 2) //> there are 9 2-permutations with replacement for abc = List(aa, ab, ac, ba,
//| bb, bc, ca, cb, cc)
noReplacements("abc", 2) //> there are 6 2-permutations without replacement for Queue(a, b, c) = Queue(b
//| a, ca, cb, ab, ac, bc)
noReplacements("abc", 3) //> there are 6 3-permutations without replacement for Queue(a, b, c) = Queue(c
//| ba, bca, acb, cab, bac, abc)
withReplacements("abc", 3) //> there are 27 3-permutations with replacement for abc = List(aaa, aab, aac,
//| aba, abb, abc, aca, acb, acc, baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc,
//| caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc)
// you can run with replacements (3 chars, n = 4) but noReplacements will fail for obvious reason -- you cannont combine 3 chars to produce 4
withReplacements("abc", 4) //> there are 81 4-permutations with replacement for abc = List(aaaa, aaab, aaa
//| c, aaba, aabb, aabc, aaca, aacb, aacc, abaa, abab, abac, abba, abbb, abbc,
//| abca, abcb, abcc, acaa, acab, acac, acba, acbb, acbc, acca, accb, accc, baa
//| a, baab, baac, baba, babb, babc, baca, bacb, bacc, bbaa, bbab, bbac, bbba,
//| bbbb, bbbc, bbca, bbcb, bbcc, bcaa, bcab, bcac, bcba, bcbb, bcbc, bcca, bcc
//| b, bccc, caaa, caab, caac, caba, cabb, cabc, caca, cacb, cacc, cbaa, cbab,
//| cbac, cbba, cbbb, cbbc, cbca, cbcb, cbcc, ccaa, ccab, ccac, ccba, ccbb, ccb
//| c, ccca, cccb, cccc)
(1 to 3) foreach (u => noReplacements("aab", u))//> there are 3 1-permutations without replacement for Queue(a, a, b) = Queue(a
//| , a, b)
//| there are 6 2-permutations without replacement for Queue(a, a, b) = Queue(a
//| a, ba, ba, aa, ab, ab)
//| there are 6 3-permutations without replacement for Queue(a, a, b) = Queue(b
//| aa, aba, aba, baa, aab, aab)
這些是相同的3行代碼，但可變長度排列，支持和列表級聯被淘汰更好。
我已經提出了更多的意識形態（這樣可以防止累加器的平面圖合併，這也使得它更具尾部遞歸性）並且擴展爲多重集合排列，以便您可以說「aab」，「aba 「和」咩「是相互排列的。這個想法是，字母「a」可以無限次（可替換的情況下）可替換兩次或僅可用一次（無替換）。所以，你需要一個計數器，它告訴你每個字母有多少次可供替換。
// Rewrite with replacement a bit to eliminate flat-map merges.
def norep2(chars: String, n: Int/* = chars.length*/) {
import scala.collection.immutable.Queue
type Set = Queue[Char]
val set = Queue[Char](chars.toList: _*)
type Result = Queue[String]
def siblings(set: (Char, Set), offset: Int, path: String, acc: Result): Result = set match {case (bubble, queue) =>
val children = descend(queue, path + bubble, acc) // bubble was used, it is not available for children that will produce combinations in other positions
if (offset == 0) children else siblings(queue.enqueue(bubble).dequeue, offset - 1, path, children) // siblings will produce different chars at the same position, fetch next char for them
}
def descend(set: Set, path: String, acc: Result): Result = {
if (path.length == n) acc.enqueue(path) else siblings(set.dequeue, set.size-1, path, acc)
}
val res = descend(set, "", Queue.empty)
println("there are " + res.length + " " + n + "-permutations without replacement for " + set + " = " + res)
} //> norep2: (chars: String, n: Int)Unit
assert(norep2("abc", 2) == noReplacements("abc", 2))
//> there are 6 2-permutations without replacement for Queue(a, b, c) = Queue(a
//| b, ac, bc, ba, ca, cb)
//| there are 6 2-permutations without replacement for Queue(a, b, c) = Queue(b
//| a, ca, cb, ab, ac, bc)
assert(norep2("abc", 3) == noReplacements("abc", 3))
//> there are 6 3-permutations without replacement for Queue(a, b, c) = Queue(a
//| bc, acb, bca, bac, cab, cba)
//| there are 6 3-permutations without replacement for Queue(a, b, c) = Queue(c
//| ba, bca, acb, cab, bac, abc)
def multisets(chars: String, n: Int/* = chars.length*/) {
import scala.collection.immutable.Queue
type Set = Queue[Bubble]
type Bubble = (Char, Int)
type Result = Queue[String]
def siblings(set: (Bubble, Set), offset: Int, path: String, acc: Result): Result = set match {case ((char, avail), queue) =>
val children = descend(if (avail - 1 == 0) queue else queue.enqueue(char -> {avail-1}), path + char, acc) // childern can reuse the symbol while if it is available
if (offset == 0) children else siblings(queue.enqueue((char, avail)).dequeue, offset - 1, path, children)
}
def descend(set: Set, path: String, acc: Result): Result = {
if (path.length == n) acc.enqueue(path) else siblings(set.dequeue, set.size-1, path, acc)
}
val set = Queue[Bubble]((chars.toList groupBy (c => c) map {case (k, v) => (k, v.length)}).toList: _*)
val res = descend(set, "", Queue.empty)
println("there are " + res.length + " multiset " + n + "-permutations for " + set + " = " + res)
} //> multisets: (chars: String, n: Int)Unit
assert(multisets("abc", 2) == norep2("abc", 2)) //> there are 6 multiset 2-permutations for Queue((b,1), (a,1), (c,1)) = Queue(
//| ba, bc, ac, ab, cb, ca)
//| there are 6 2-permutations without replacement for Queue(a, b, c) = Queue(a
//| b, ac, bc, ba, ca, cb)
assert(multisets("abc", 3) == norep2("abc", 3)) //> there are 6 multiset 3-permutations for Queue((b,1), (a,1), (c,1)) = Queue(
//| bac, bca, acb, abc, cba, cab)
//| there are 6 3-permutations without replacement for Queue(a, b, c) = Queue(a
//| bc, acb, bca, bac, cab, cba)
assert (multisets("aaab", 2) == multisets2("aaab".toList, 2))
//> there are 3 multiset 2-permutations for Queue((b,1), (a,3)) = Queue(ba, ab,
//| aa)
//| there are 3 multiset 2-permutations for Queue((b,1), (a,3)) = List(List(a,
//| a), List(b, a), List(a, b))
multisets("aab", 2) //> there are 3 multiset 2-permutations for Queue((b,1), (a,2)) = Queue(ba, ab,
//| aa)
multisets("aab", 3) //> there are 3 multiset 3-permutations for Queue((b,1), (a,2)) = Queue(baa, ab
//| a, aab)
norep2("aab", 3) //> there are 6 3-permutations without replacement for Queue(a, a, b) = Queue(a
//| ab, aba, aba, aab, baa, baa)
由於generalizaiton，你可以用/獲取，而無需使用多集功能可按替代品。例如，
//take far more letters than resulting permutation length to emulate withReplacements
assert(multisets("aaaaabbbbbccccc", 3) == withReplacements("abc", 3))
//> there are 27 multiset 3-permutations for Queue((b,5), (a,5), (c,5)) = Queue
//| (bac, bab, baa, bcb, bca, bcc, bba, bbc, bbb, acb, aca, acc, aba, abc, abb,
//| aac, aab, aaa, cba, cbc, cbb, cac, cab, caa, ccb, cca, ccc)
//| there are 27 3-permutations with replacement for abc = List(aaa, aab, aac,
//| aba, abb, abc, aca, acb, acc, baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc,
//| caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc)
//take one letter of each to emulate withoutReplacements
assert(multisets("aaaaabbbbbccccc", 3) == noReplacements("abc", 3))
//> there are 27 multiset 3-permutations for Queue((b,5), (a,5), (c,5)) = Queue
//| (bac, bab, baa, bcb, bca, bcc, bba, bbc, bbb, acb, aca, acc, aba, abc, abb,
//| aac, aab, aaa, cba, cbc, cbb, cac, cab, caa, ccb, cca, ccc)
//| there are 6 3-permutations without replacement for Queue(a, b, c) = Queue(c
//| ba, bca, acb, cab, bac, abc)
If you are more interested about permutations, you may look at
這是基於span版本。
def perms[T](xs: List[T]): List[List[T]] = xs match {
case List(_) => List(xs)
case _ => for (x <- xs
; val (l, r) = xs span { x!= }
; ys <- perms(l ++ r.tail)
) yield x :: ys
}
def permutator[T](list: List[T]): List[List[T]] = {
def _p(total: Int, list: List[T]): List[List[T]] = {
if (total == 0) {
// End of the recursion tree
list.map(List(_))
} else {
// Controlled combinatorial
// explosion while total > 0
for (i <- list;
j <- _p(total - 1, list))
yield { i :: j }
// It is a recursion tree to generate the
// permutations of the elements
// --------------------------------------
// total = 0 => _p returns 3 elements (A, B, C)
// total = 1 => _p returns 3 * 3 List(List(A, A)...
// total = 2 => _p returns 3 * 3 * 3 elements List(List(A, A, A)...
}
}
_p(list.length - 1, list)
}
permutator(List("A", "B", "C"))
// output:
List(A, A, A),List(A, A, B),List(A, A, C),List(A, B, A),List(A, B, B),
List(A, B, C),List(A, C, A),List(A, C, B),List(A, C, C),List(B, A, A),
List(B, A, B),List(B, A, C),List(B, B, A),List(B, B, B),List(B, B, C),
List(B, C, A),List(B, C, B),List(B, C, C),List(C, A, A),List(C, A, B),
List(C, A, C),List(C, B, A),List(C, B, B),List(C, B, C),List(C, C, A),
List(C, C, B),List(C, C, C)
請解釋答案，代碼只回答不是很有幫助。 – Jeet
這是基於週期的概念和一個簡單的實現置換的具有兩個元件的實現方式。 它不處理重複和堆棧溢出方面的置換方法
object ImmuPermute extends App {
def nextCycle(nums: List[Int]): List[Int] = {
nums match {
case Nil => Nil
case head :: tail => tail :+ head
}
}
def cycles(nums: List[Int]): List[List[Int]] = {
def loop(l: List[Int], acc: List[List[Int]]): List[List[Int]] = {
if (acc.size == nums.size)
acc
else {
val next = nextCycle(l)
loop(next, next :: acc)
}
}
loop(nums, List(nums))
}
def permute(nums: List[Int]): List[List[Int]] = {
nums match {
case Nil => Nil
case head :: Nil => List(List(head))
case first :: second :: Nil => List(List(first, second), List(second, first))
case _ => {
val cycledList = cycles(nums)
cycledList.flatMap { cycle =>
val h = cycle.head
val t = cycle.tail
val permutedList = permute(t)
permutedList map { pList =>
h :: pList
}
}
}
}
}
val l = permute(List(1, 2, 3, 4))
l foreach println
println(l.size)
}
這也許應該是codereview.SE。 – Raphael
@Raphael我不得不穀歌那一個，所以這裏是懶惰http://codereview.stackexchange.com/ – simbo1905
我認爲OP是好的SO。人們需要了解其他人可能會如何處理某些問題，在這裏進行排列，以改進Scala編程。 – lcn