Ios 在Swift中将对象数组排序为每个可能的序列
想知道在Swift中是否有一种干净的方法可以做到这一点。可能使用一个或几个全局函数,如Map/Reduce等 数组包含数量为n的唯一自定义对象 例如,有3个项目。但可能会有更多或更少[1,2,3] 将返回一个数组数组 [ [1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1] ] 完成任务。只需要进入Swift状态。 他创建了结构来处理排列:Ios 在Swift中将对象数组排序为每个可能的序列,ios,arrays,swift,sorting,Ios,Arrays,Swift,Sorting,想知道在Swift中是否有一种干净的方法可以做到这一点。可能使用一个或几个全局函数,如Map/Reduce等 数组包含数量为n的唯一自定义对象 例如,有3个项目。但可能会有更多或更少[1,2,3] 将返回一个数组数组 [ [1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1] ] 完成任务。只需要进入Swift状态。 他创建了结构来处理排列: var greetingPermutations = PermutationSequenc
var greetingPermutations = PermutationSequenceGenerator(elements: ["hi", "hey", "hello"])
while let greetingSequence = greetingPermutations.next(){
for greeting in greetingSequence {
print("\(greeting) ")
}
println()
}
或:
编辑:
下面是在objc.io上找到的一种更简单的方法
添加扩展名
extension Array {
var decompose : (head: T, tail: [T])? {
return (count > 0) ? (self[0], Array(self[1..<count])) : nil
}
}
这段代码使用分解函数扩展了数组,还添加了>>==操作符(展平)更多关于展平的内容:可能太过c-ish,但这里是已经发布的示例的替代方案
var a = [1, 2, 3, 4, 5]
var b = [[Int]]()
func perms<T>(n: Int, inout a: [T], inout b: [[T]]) {
if n == 0 {
b.append(a)
} else {
for i in 0..<n {
perms(n - 1, &a, &b)
var j = 0
if n % 2 == 0 {
j = i
}
swap(&a[j], &a[n - 1])
}
}
}
perms(a.count, &a, &b)
println(b)
var a=[1,2,3,4,5]
变量b=[[Int]]()
函数perms(n:Int,inout a:[T],inout b:[T]]{
如果n==0{
b、 附加(a)
}否则{
对于0中的i..Swift 5
更新版本的@DogCoffee for swift 5.x,均在阵列扩展中:
extension Array {
private var decompose : (head: Element, tail: [Element])? {
return (count > 0) ? (self[0], Array(self[1..<count])) : nil
}
private func between<T>(x: T, ys: [T]) -> [[T]] {
if let (head, tail) = ys.decompose {
return [[x] + ys] + between(x: x, ys: tail).map { [head] + $0 }
} else {
return [[x]]
}
}
private func permutations<T>(xs: [T]) -> [[T]] {
if let (head, tail) = xs.decompose {
return permutations(xs: tail) >>= { permTail in
self.between(x: head, ys: permTail)
}
} else {
return [[]]
}
}
func allPermutations() -> [[Element]] {
return permutations(xs: self)
}
}
infix operator >>=
func >>=<A, B>(xs: [A], f: (A) -> [B]) -> [B] {
return xs.map(f).reduce([], +)
}
扩展数组{
私有var分解:(head:Element,tail:[Element]){
返回(计数>0)?(自[0],数组(自[1..[[T]]{
如果让(头、尾)=ys分解{
返回[[x]+ys]+between(x:x,ys:tail).map{[head]+$0}
}否则{
返回[[x]]
}
}
私有函数置换(xs:[T])->[[T]]{
如果let(head,tail)=xs.decompose{
返回置换(xs:tail)>>={permTail in
self.between(x:head,ys:permTail)
}
}否则{
返回[]]
}
}
func allPermutations()->[[Element]]{
返回置换(xs:self)
}
}
中缀运算符>>=
func>>=(xs[A],f:(A)->[B])->[B]{
返回xs.map(f).reduce([],+)
}
what is between in:between(head,permTail)对不起,忘了粘贴between函数
func between<T>(x: T, ys: [T]) -> [[T]] {
if let (head, tail) = ys.decompose {
return [[x] + ys] + between(x, ys: tail).map { [head] + $0 }
} else {
return [[x]]
}
}
func permutations<T>(xs: [T]) -> [[T]] {
if let (head, tail) = xs.decompose {
return permutations(tail) >>= { permTail in
self.between(head, ys: permTail)
}
} else {
return [[]]
}
}
let example = permutations([1,2,3,5,6,7,8])
println(example)
var a = [1, 2, 3, 4, 5]
var b = [[Int]]()
func perms<T>(n: Int, inout a: [T], inout b: [[T]]) {
if n == 0 {
b.append(a)
} else {
for i in 0..<n {
perms(n - 1, &a, &b)
var j = 0
if n % 2 == 0 {
j = i
}
swap(&a[j], &a[n - 1])
}
}
}
perms(a.count, &a, &b)
println(b)
extension Array {
private var decompose : (head: Element, tail: [Element])? {
return (count > 0) ? (self[0], Array(self[1..<count])) : nil
}
private func between<T>(x: T, ys: [T]) -> [[T]] {
if let (head, tail) = ys.decompose {
return [[x] + ys] + between(x: x, ys: tail).map { [head] + $0 }
} else {
return [[x]]
}
}
private func permutations<T>(xs: [T]) -> [[T]] {
if let (head, tail) = xs.decompose {
return permutations(xs: tail) >>= { permTail in
self.between(x: head, ys: permTail)
}
} else {
return [[]]
}
}
func allPermutations() -> [[Element]] {
return permutations(xs: self)
}
}
infix operator >>=
func >>=<A, B>(xs: [A], f: (A) -> [B]) -> [B] {
return xs.map(f).reduce([], +)
}