我已经使用嵌套循环在scala中编写了一种方法pythagoreanTriplets
。作为scala中的新手,我正在努力使用递归如何将返回列表(元组列表)使用懒惰求值方法。任何帮助将不胜感激。
P.S:以下方法运行良好。
// This method returns the list of all pythagorean triples whose components are
// at most a given limit. Formula a^2 + b^2 = c^2
def pythagoreanTriplets(limit: Int): List[(Int, Int, Int)] = {
// triplet: a^2 + b^2 = c^2
var (a,b,c,m) = (0,0,0,2)
var triplets:List[(Int, Int, Int)] = List()
while (c < limit) {
breakable {
for (n <- 1 until m) {
a = m * m - n * n
b = 2 * m * n
c = m * m + n * n
if (c > limit)
break
triplets = triplets :+ (a, b, c)
}
m += 1
}
}// end of while
triplets
}
我看不到递归将在哪些方面提供明显的优势。
def pythagoreanTriplets(limit: Int): List[(Int, Int, Int)] =
for {
m <- (2 to limit/2).toList
n <- 1 until m
c = m*m + n*n if c <= limit
} yield (m*m - n*n, 2*m*n, c)