合并重叠的字符串答案

【问题标题】：Merging overlapping strings合并重叠的字符串
【发布时间】：2020-04-19 18:03:00
【问题描述】：

假设我需要像这样合并两个重叠的字符串：

def mergeOverlap(s1: String, s2: String): String = ???

mergeOverlap("", "")       // ""
mergeOverlap("", "abc")    // abc  
mergeOverlap("xyz", "abc") // xyzabc 
mergeOverlap("xab", "abc") // xabc

我可以使用answer 为我之前的一个问题编写此函数：

def mergeOverlap(s1: String, s2: String): String = { 
  val n = s1.tails.find(tail => s2.startsWith(tail)).map(_.size).getOrElse(0)
  s1 ++ s2.drop(n)
}

您能否建议或者一个更简单的或可能更有效的mergeOverlap实现？

【问题讨论】：

标签： string scala collections

【解决方案1】：

您可以使用算法计算prefix function，在与字符串总长度成正比的时间上找到两个字符串之间的重叠O(n + k)。索引i处字符串的前缀函数定义为索引i处最长后缀的大小，等于整个字符串的前缀（不包括平凡的情况）。

有关定义和计算它的算法的更多解释，请参阅这些链接：

这里是一个修改算法的实现，它计算第二个参数的最长前缀，等于第一个参数的后缀：

import scala.collection.mutable.ArrayBuffer

def overlap(hasSuffix: String, hasPrefix: String): Int = {
  val overlaps = ArrayBuffer(0)
  for (suffixIndex <- hasSuffix.indices) {
    val currentCharacter = hasSuffix(suffixIndex)
    val currentOverlap = Iterator.iterate(overlaps.last)(overlap => overlaps(overlap - 1))
      .find(overlap =>
        overlap == 0 ||
        hasPrefix.lift(overlap).contains(currentCharacter))
      .getOrElse(0)
    val updatedOverlap = currentOverlap +
      (if (hasPrefix.lift(currentOverlap).contains(currentCharacter)) 1 else 0)
    overlaps += updatedOverlap
  }
  overlaps.last
}

而mergeOverlap 只是

def mergeOverlap(s1: String, s2: String) = 
  s1 ++ s2.drop(overlap(s1, s2))

以及对这个实现的一些测试：

scala> mergeOverlap("", "")    
res0: String = ""

scala> mergeOverlap("abc", "")
res1: String = abc

scala> mergeOverlap("", "abc")
res2: String = abc

scala> mergeOverlap("xyz", "abc")
res3: String = xyzabc

scala> mergeOverlap("xab", "abc")
res4: String = xabc

scala> mergeOverlap("aabaaab", "aab")
res5: String = aabaaab

scala> mergeOverlap("aabaaab", "aabc")
res6: String = aabaaabc

scala> mergeOverlap("aabaaab", "bc")
res7: String = aabaaabc

scala> mergeOverlap("aabaaab", "bbc")
res8: String = aabaaabbc

scala> mergeOverlap("ababab", "ababc")
res9: String = abababc

scala> mergeOverlap("ababab", "babc")
res10: String = abababc

scala> mergeOverlap("abab", "aab")
res11: String = ababaab

【讨论】：

非常感谢。解决方案并不简单，但绝对非常有效。看起来很有趣。我将学习这个算法并尝试自己实现它。
不幸的是我很难理解updatedOverlap的计算方法:((问了一个新问题(stackoverflow.com/questions/63923861/…)。

【解决方案2】：

它不是尾递归，但它是一个非常简单的算法。

def mergeOverlap(s1: String, s2: String): String =
  if (s2 startsWith s1) s2
  else s1.head +: mergeOverlap(s1.tail, s2)

【讨论】：

是的，但是效率很低（这是 OP 要求的）。
我询问了更简单或更高效的实现。将更新问题以强调它。