Scala 如何将序列拆分为子序列?
假设我有一个整数序列,需要将其拆分为如下子序列:Scala 如何将序列拆分为子序列?,scala,collections,Scala,Collections,假设我有一个整数序列,需要将其拆分为如下子序列: def splitBySeq(xs: Seq[Int], ys: Seq[Int]): (Seq[Int], Seq[Int]) = ??? val xs = List(1, 2, 3, 4, 5) splitBySeq(xs, Nil) // (List(1, 2, 3, 4, 5), Nil) splitBySeq(xs, List(1)) // (Nil, List(2, 3, 4,
def splitBySeq(xs: Seq[Int], ys: Seq[Int]): (Seq[Int], Seq[Int]) = ???
val xs = List(1, 2, 3, 4, 5)
splitBySeq(xs, Nil) // (List(1, 2, 3, 4, 5), Nil)
splitBySeq(xs, List(1)) // (Nil, List(2, 3, 4, 5))
splitBySeq(xs, List(5)) // (List(1, 2, 3, 4), Nil)
splitBySeq(xs, List(3, 4)) // (List(1, 2), List(5))
splitBySeq(xs, List(11, 12)) // (List(1, 2, 3, 4, 5), Nil)
splitBySeq(xs, List(1, 2, 3, 4, 5)) // (Nil, Nil)
如果ys是xs的子序列,那么函数应该返回一对序列-xs1和xs2,这样xs1++ys++xs2==xs。否则,函数返回xs
您将如何实现splitBySeq?这似乎说明了您的目标
def splitBySeq(xs: Seq[Int], ys: Seq[Int]): (Seq[Int], Seq[Int]) = {
val idx = xs indexOfSlice ys
if (idx < 0) (xs, Nil)
else {
val (a,b) = xs splitAt idx
(a, b drop ys.length)
}
}
请注意,在第一个测试用例splitBySeqxs,Nil中,结果seqs被切换,因为Nil匹配任何Seq的零索引。这似乎是您所追求的
def splitBySeq(xs: Seq[Int], ys: Seq[Int]): (Seq[Int], Seq[Int]) = {
val idx = xs indexOfSlice ys
if (idx < 0) (xs, Nil)
else {
val (a,b) = xs splitAt idx
(a, b drop ys.length)
}
}
注意,在第一个测试用例splitBySeqxs,Nil中,结果seqs被切换,因为Nil匹配任何Seq的零索引。另一个解决方案,尾部递归函数进行一次迭代:
def splitBySeq[A](xs: Seq[A], ys: Seq[A]): (Seq[A], Seq[A]) = {
@tailrec
def go(a: List[A], b: List[A], acc: List[A], rest: List[A]): (Seq[A], Seq[A]) = {
(a, b) match {
case (z :: zs, w :: ws) => {
if(z == w) {
go(zs, ws, z :: acc, rest)
} else{
go(zs, ys.toList, Nil, z :: (acc ++ rest))
}
}
case (zs, Nil) => (rest.reverse, zs)
case (Nil, _) => (rest.reverse, Nil)
}
}
if(ys.isEmpty) {
(xs, Nil)
} else {
go(xs.toList, ys.toList, Nil, Nil)
}
}
另一种解决方案,尾部递归函数执行一次迭代:
def splitBySeq[A](xs: Seq[A], ys: Seq[A]): (Seq[A], Seq[A]) = {
@tailrec
def go(a: List[A], b: List[A], acc: List[A], rest: List[A]): (Seq[A], Seq[A]) = {
(a, b) match {
case (z :: zs, w :: ws) => {
if(z == w) {
go(zs, ws, z :: acc, rest)
} else{
go(zs, ys.toList, Nil, z :: (acc ++ rest))
}
}
case (zs, Nil) => (rest.reverse, zs)
case (Nil, _) => (rest.reverse, Nil)
}
}
if(ys.isEmpty) {
(xs, Nil)
} else {
go(xs.toList, ys.toList, Nil, Nil)
}
}
splitBySeqxs,List5和splitBySeqxs,List3,4的示例结果与您的描述不匹配。同意。我会修好的。谢谢。splitBySeqxs,List5和splitBySeqxs,List3,4的示例结果与您的描述不匹配。同意。我会修好的。谢谢