问使用元组从递归函数中提取数据

EN

如果返回类型固定为List[List[String]]，则需要对代码进行以下更改：

1. 因为someType._2是作为someType._2(2)访问的，所以someType._2列表中至少应该有3字符串。
2. 最后一个表达式必须是返回类型ie，List[List[String]]。因为someType._2(1)和someType._2(2)只是字符串而不是List[String]：List(someTuple._2,List(someTuple._2(1),someTuple._2(2)))将是返回类型，值"Some Word“将在递归过程中改变，注意到someTuple._2.size总是>=3.
3. As我们需要访问someType._2，它将在每次递归过程中改变，它在递归函数中被声明为var。

有了从您的需求中得出的理解，下面的代码可能就是您正在寻找的代码：

def someRecursiveFunction(listOfWords:ListString，sw: String):List[ListString] ={ val textSplitter = listOfWords.lastIndexOf(sw) var i =0 if(i==0) { var someTuple:(ListString，ListString) = (List()，List()) } if (textSplitter != -1 && listOfWords.size-3>=textSplitter) { someTuple = listOfWords.splitAt(textSplitter) println(someTuple._1，someTuple._2) //用于检查递归if( someTuple._1.size>=3){ i+=1 someRecursiveFunction(someTuple._1，someTuple._1(textSplitter-3))} } List(someTuple._2，List(someTuple._2(1)，someTuple._2(2) // What I want back }

Scala中的：

val list = List("a","b","c","x","y","z","k","j","g","Some Word","d","e","f","u","m","p")


scala> val list = List("a","b","c","x","y","z","k","j","g","Some Word","d","e","f","u","m","p")
list: List[String] = List(a, b, c, x, y, z, k, j, g, Some Word, d, e, f, u, m, p)

scala> someRecursiveFunction(list,"d")
(List(a, b, c, x, y, z, k, j, g, Some Word),List(d, e, f, u, m, p))
(List(a, b, c, x, y, z, k),List(j, g, Some Word))
(List(a, b, c, x),List(y, z, k))
(List(a),List(b, c, x))
res70: List[List[String]] = List(List(b, c, x), List(c, x))

scala> someRecursiveFunction(list,"Some Word")
(List(a, b, c, x, y, z, k, j, g),List(Some Word, d, e, f, u, m, p))
(List(a, b, c, x, y, z),List(k, j, g))
(List(a, b, c),List(x, y, z))
(List(),List(a, b, c))
res71: List[List[String]] = List(List(a, b, c), List(b, c))