使用Scala查找给定字符串多少次是另一个字符串的子字符串

Question

用scala查找给定字符串多少次是另一个字符串的子字符串的优雅方法是什么？

以下测试用例应明确要求是什么：

import org.scalatest.FunSuite

class WordOccurrencesSolverTest extends FunSuite {

  private val solver = new WordOccurrencesSolver()

  test("solve for a in a") {
    assert(solver.solve("a", "a") === 1)
  }

  test("solve for b in a") {
    assert(solver.solve("b", "a") === 0)
  }

  test("solve for a in aa") {
    assert(solver.solve("a", "aa") === 2)
  }

  test("solve for b in ab") {
    assert(solver.solve("b", "ab") === 1)
  }

  test("solve for ab in ab") {
    assert(solver.solve("ab", "ab") === 1)
  }

  test("solve for ab in abab") {
    assert(solver.solve("ab", "abab") === 2)
  }

  test("solve for aa in aaa") {
    assert(solver.solve("aa", "aaa") === 2)
  }
}

这是我对此并不特别自豪的解决方案：

class WordOccurrencesSolver {

  def solve(word: String, text: String): Int = {
    val n = word.length
    def solve(acc: Int, word: String, sb: String): Int = sb match {
      case _ if sb.length < n => acc
      case _ if sb.substring(0, n) == word => solve(acc + 1, word, sb.tail)
      case _ => solve(acc, word, sb.tail)
    }
    solve(0, word, text)
  }

}

我假设必须有一个干净的衬板，该衬板利用Scala的高阶函数而不是递归和match / case子句。

Answer 1

如果您正在寻找惯用的Scala解决方案，则可以使用sliding创建一个滑动窗口迭代器，并计算与目标String相等的窗口数。

该解决方案在发挥功能的同时，还可以为您提供令人满意的性能。

def countOccurrences(src: String, tgt: String): Int =
  src.sliding(tgt.length).count(window => window == tgt)