查找数组的最小和最大元素

这是一个简洁易读的解决方案,可以避免丑陋的if陈述:

def minMax(a: Array[Int]) : (Int, Int) = {
  if (a.isEmpty) throw new java.lang.UnsupportedOperationException("array is empty")
  a.foldLeft((a(0), a(0)))
  { case ((min, max), e) => (math.min(min, e), math.max(max, e))}
}

Explanation:foldLeftScala中许多集合的标准方法.它允许将累加器传递给将为数组的每个元素调用的回调函数.

请查看scaladoc以获取更多详细信息

遵循其他答案-可能有一个更通用的解决方案，该解决方案适用于其他集合Array，和其他内容以及Int：

def minmax[B >: A, A](xs: Iterable[A])(implicit cmp: Ordering[B]): (A, A) = {
  if (xs.isEmpty) throw new UnsupportedOperationException("empty.minmax")

  val initial = (xs.head, xs.head)

  xs.foldLeft(initial) { case ((min, max), x) => 
    (if (cmp.lt(x, min)) x else min, if (cmp.gt(x, max)) x else max) }
}                                               

例如：

minmax(List(4, 3, 1, 2, 5))              //> res0: (Int, Int) = (1,5)

minmax(Vector('Z', 'K', 'B', 'A'))       //> res1: (Char, Char) = (A,Z)

minmax(Array(3.0, 2.0, 1.0))             //> res2: (Double, Double) = (1.0,3.0)

（也可以使用cmp.min()和来更简洁地编写此代码cmp.max()，但前提是要删除B >: A类型绑定，这会使函数的通用性降低）。

val xs: Array[Int] = ???

var min: Int = Int.MaxValue
var max: Int = Int.MinValue

for (x <- xs) {
  if (x < min) min = x
  if (x > max) max = x
}

def findMinAndMax(array: Array[Int]) = { // a non-empty array

    val initial = (array.head, array.head)  // a tuple representing min-max


    // foldLeft takes an initial value of type of result, in this case a tuple
    // foldLeft also takes a function of 2 parameters.
    // the 'left' parameter is an accumulator (foldLeft -> accum is left)
    // the other parameter is a value from the collection.

    // the function2 should return a value which replaces accumulator (in next iteration)
    // when the next value from collection will be picked.

    // so on till all values are iterated, in the end accum is returned.

    array.foldLeft(initial) { ((min, max), x) => 
          if (x < min) (x, max) 
          else if (x > max) (min, x) 
          else acc 
    }
}

考虑一下(对于非空的可订购数组),

val ys = xs.sorted
val (min,max) = (ys.head, ys.last)

您可以使用reduceLeft函数获取Array [Int]的最小值和最大值.

scala> val a = Array(20, 12, 6, 15, 2, 9)

a:Array [Int] = Array(20,12,6,15,2,9)

scala> a.reduceLeft(_ min _)

res:Int = 2

scala> a.reduceLeft(_ max _)

res:Int = 20

有关更多信息和reduceLeft方法示例,请参阅此链接:http://alvinalexander.com/scala/scala-reduceleft-examples