斯卡拉：带密封盖

Question

重写代码重复我有这样的代码：

val arr: Array[Int] = ... 
val largestIndex = { 
    var i = arr.length-2 
    while (arr(i) > arr(i+1)) i -= 1 
    i 
} 
val smallestIndex = { 
    var k = arr.length-1 
    while (arr(largestIndex) > arr(k)) k -= 1 
    k 
} 

但有得多代码重复。我试图用闭包重写这个，但是我失败了。我试过这样的事情：

def index(sub: Int, f: => Boolean): Int = { 
    var i = arr.length-sub 
    while (f) i -= 1 
    i 
} 
val largest = index(2, i => arr(i) > arr(i+1)) 
val smallest = index(1, i => arr(largest) > arr(i)) 

问题是我不能在闭包中使用方法index（）的参数i。有没有办法避免这个问题？

Answer 1

val arr = Array(1,2,4,3,3,4,5) 
def index(sub: Int, f: Int => Boolean): Int = { 
    var i = arr.length-sub      
    while (f(i)) i -= 1       
    i            
}            
val largest = index(2, i => arr(i) > arr(i+1)) 
val smallest = index(1, i => arr(largest) > arr(i)) 

Answer 2

val arr = Array(1,2,4,3,3,4,5) 
arr: Array[Int] = Array(1, 2, 4, 3, 3, 4, 5) 

scala> arr.zipWithIndex.max(Ordering.by((x: (Int, Int)) => x._1))._2 
res0: Int = 6 

scala> arr.zipWithIndex.min(Ordering.by((x: (Int, Int)) => x._1))._2 
res1: Int = 0 

或

scala> val pairOrdering = Ordering.by((x: (Int, Int)) => x._1) 
pairOrdering: scala.math.Ordering[(Int, Int)] = scala.math.Ordering$$anon$4@145ad3d 

scala> arr.zipWithIndex.max(pairOrdering)._2 
res2: Int = 6 

scala> arr.zipWithIndex.min(pairOrdering)._2 
res3: Int = 0