springlustre的专栏

Implement the so-called run-length encoding data compression method directly. I.e. don't use other methods you've written (like P09's pack); do all the work directly.Example:scala> encodeDirect(

Example:scala> range(4, 9)res0: List[Int] = List(4, 5, 6, 7, 8, 9)def range(n1:Int,n2:Int)={ List(n1 to n2)}看了下答案，原来可以直接用List.range()递归解法def range(start: Int, end: Int): List[Int] = if (end <

Return the list and the removed element in a Tuple. Elements are numbered from 0.Example:scala> removeAt(1, List('a, 'b, 'c, 'd))res0: (List[Symbol], Symbol) = (List('a, 'c, 'd),'b)def removeAt

Modify the result of problem P10 in such a way that if an element has no duplicates it is simply copied into the result list. Only elements with duplicates are transferred as (N, E) terms.Example:

16scala> drop(3, List('a, 'b, 'c, 'd, 'e, 'f, 'g, 'h, 'i, 'j, 'k))res0: List[Symbol] = List('a, 'b, 'd, 'e, 'g, 'h, 'j, 'k)参考答案：其中关于zip 和filter的用法见http://blog.youkuaiyun.com/springlustre/article/deta

Given a run-length code list generated as specified in problem P10, construct its uncompressed version.Example:scala> decode(List((4, 'a), (1, 'b), (2, 'c), (2, 'a), (1, 'd), (4, 'e)))res0: Lis

Examples:scala> rotate(3, List('a, 'b, 'c, 'd, 'e, 'f, 'g, 'h, 'i, 'j, 'k))res0: List[Symbol] = List('d, 'e, 'f, 'g, 'h, 'i, 'j, 'k, 'a, 'b, 'c)scala> rotate(-2, List('a, 'b, 'c, 'd, 'e, 'f, 'g,

scala> duplicate(List('a, 'b, 'c, 'c, 'd))res0: List[Symbol] = List('a, 'a, 'b, 'b, 'c, 'c, 'c, 'c, 'd, 'd)scala> duplicateN(3, List('a, 'b, 'c, 'c, 'd))res0: List[Symbol] = List('a, 'a, 'a, 'b, 'b,

Given two indices, I and K, the slice is the list containing the elements from and including the Ith element up to but not including the Kth element of the original list. Start counting the elemen

Example:scala> last(List(1, 1, 2, 3, 5, 8))res0: Int = 8最直接的办法def s1(list:List[Any])={ list.last}def main(args: Array[String]) { println(s1(List(1,2,3,4,5)))}