所以,在回答其他问题时,我偶然发现了计算中位数 5 的必要性。现在,另一种语言也有类似的问题,但我想要一个 Scala 算法,我不确定我是否满意.
问问题
1981 次
3 回答
5
这是一个不可变的 Scala 版本,它的比较次数最少 (6),而且看起来不太难看:
def med5(five: (Int,Int,Int,Int,Int)) = {
// Return a sorted tuple (one compare)
def order(a: Int, b: Int) = if (a<b) (a,b) else (b,a)
// Given two self-sorted pairs, pick the 2nd of 4 (two compares)
def pairs(p: (Int,Int), q: (Int,Int)) = {
(if (p._1 < q._1) order(p._2,q._1) else order(q._2,p._1))._1
}
// Strategy is to throw away smallest or second smallest, leaving two self-sorted pairs
val ltwo = order(five._1,five._2)
val rtwo = order(five._4,five._5)
if (ltwo._1 < rtwo._1) pairs(rtwo,order(ltwo._2,five._3))
else pairs(ltwo,order(rtwo._2,five._3))
}
编辑:按照 Daniel 的要求,这是一个适用于所有尺寸和数组的修改,因此它应该是有效的。我不能让它漂亮,所以效率是次要的。(>200M 中位数/秒,预分配数组为 5,比 Daniel 的版本快 100 多倍,比我上面的不可变版本快 8 倍(长度为 5))。
def med5b(five: Array[Int]): Int = {
def order2(a: Array[Int], i: Int, j: Int) = {
if (a(i)>a(j)) { val t = a(i); a(i) = a(j); a(j) = t }
}
def pairs(a: Array[Int], i: Int, j: Int, k: Int, l: Int) = {
if (a(i)<a(k)) { order2(a,j,k); a(j) }
else { order2(a,i,l); a(i) }
}
if (five.length < 2) return five(0)
order2(five,0,1)
if (five.length < 4) return (
if (five.length==2 || five(2) < five(0)) five(0)
else if (five(2) > five(1)) five(1)
else five(2)
)
order2(five,2,3)
if (five.length < 5) pairs(five,0,1,2,3)
else if (five(0) < five(2)) { order2(five,1,4); pairs(five,1,4,2,3) }
else { order2(five,3,4); pairs(five,0,1,3,4) }
}
于 2011-01-21T17:40:47.070 回答
2
天哪,想多了,伙计们。
def med5(a : Int, b: Int, c : Int, d : Int, e : Int) =
List(a, b, c, d, e).sort(_ > _)(2)
于 2011-01-22T06:44:58.417 回答
0
正如所建议的,这是我自己的算法:
def medianUpTo5(arr: Array[Double]): Double = {
def oneAndOrderedPair(a: Double, smaller: Double, bigger: Double): Double =
if (bigger < a) bigger
else if (a < smaller) smaller else a
def partialOrder(a: Double, b: Double, c: Double, d: Double) = {
val (s1, b1) = if (a < b) (a, b) else (b, a)
val (s2, b2) = if (c < d) (c, d) else (d, c)
(s1, b1, s2, b2)
}
def medianOf4(a: Double, b: Double, c: Double, d: Double): Double = {
val (s1, b1, s2, b2) = partialOrder(a, b, c, d)
if (b1 < b2) oneAndOrderedPair(s2, s1, b1)
else oneAndOrderedPair(s1, s2, b2)
}
arr match {
case Array(a) => a
case Array(a, b) => a min b
case Array(a, b, c) =>
if (a < b) oneAndOrderedPair(c, a, b)
else oneAndOrderedPair(c, b, a)
case Array(a, b, c, d) => medianOf4(a, b, c, d)
case Array(a, b, c, d, e) =>
val (s1, b1, s2, b2) = partialOrder(a, b, c, d)
if (s1 < s2) medianOf4(e, b1, s2, b2)
else medianOf4(e, b2, s1, b1)
}
}
于 2011-01-21T20:14:05.157 回答