我在Scala中创建了Knuth–Morris–Pratt algorithm的简单实现。现在,我想花哨并以尾递归的方式做同样的事情。我的直觉表示,应该不太困难(桌子和搜索部件都一样),但同样的感觉也告诉我,这必须已经由某个人(可能比我聪明)完成。因此是一个问题。您知道Knuth-Morris-Pratt算法的任何尾递归实现吗?
object KnuthMorrisPrattAlgorithm {
def search(s: String, w: String): Int = {
if (w.isEmpty) {
return 0
}
var m = 0
var i = 0
val t = table(w)
while(m + i < s.length) {
if (w(i) == s(m + i)) {
if (i == w.length - 1) {
return m
}
i += 1
} else {
if (t(i) > -1) {
i = t(i)
m += i - t(i)
} else {
i = 0
m += 1
}
}
}
return -1
}
def table(w: String): Seq[Int] = {
var pos = 2
var cnd = 0
val t = Array(-1, 0) ++ Array.fill(w.size - 2)(0)
while (pos < w.length) {
if (w(pos - 1) == w(cnd)) {
cnd += 1
t(pos) = cnd
pos += 1
} else if (cnd > 0) {
cnd = t(cnd)
} else {
t(pos) = 0
pos += 1
}
}
t
}
}
最佳答案
我不知道该算法的作用,但是这是您的函数,尾递归化:
object KnuthMorrisPrattAlgorithm {
def search(s: String, w: String): Int = {
if (w.isEmpty) {
return 0
}
val t = table(w)
def f(m: Int, i: Int): Int = {
if (m + i < s.length) {
if (w(i) == s(m + i)) {
if (i == w.length - 1) {
m
} else {
f(m, i + 1)
}
} else {
if (t(i) > -1) {
f(m + i - t(i), t(i))
} else {
f(m + 1, 0)
}
}
} else {
-1
}
}
f(0, 0)
}
def table(w: String): Seq[Int] = {
val t = Array(-1, 0) ++ Array.fill(w.size - 2)(0)
def f(pos: Int, cnd: Int): Array[Int] = {
if (pos < w.length) {
if (w(pos - 1) == w(cnd)) {
t(pos) = cnd + 1
f(pos + 1, cnd + 1)
} else if (cnd > 0) {
f(pos, t(cnd))
} else {
t(pos) = 0
f(pos + 1, cnd)
}
} else {
t
}
}
f(2, 0)
}
}