Question

具有查找“沙漏”和计算总和的算法

object Hourglass2D {

  def main(args: Array[String]) {

    val sc = new java.util.Scanner(System.in)
    var n = sc.nextInt()
    var arr = new Array[Int](n)
    for (arr_i <- 0 until n) {
      arr(arr_i) = sc.nextInt()
    }

    var res = -2147483648
    for (i <- 0 until arr.length - 2) {
      for (j <- 0 until arr.length - 2) {
        val sum = checkSum(arr, i, j, j + 2) + checkSum(arr,  i + 1, j + 1, j + 1) + checkSum(arr, i + 2, j, j + 2)
        if (sum > res) res = sum
      }
    }

    println(res)
  }

  private def checkSum(arr: Array[Array[Int]], row: Int, first: Int, last: Int) = {
    var sum = -1
    for (i <- first until last + 1) {
      sum = sum + arr(row)(i)
    }
    sum + 1
  }

正如你所看到的，我使用了可变状态，比如var。任何想法如何使用val而不是var？

感谢。

Answer 1

构建arr存在一些问题，所以我会跳过它。您可以使用for {...} yield表达式构建总和，然后使用max添加常量并找到最大值：

object Hourglass2D {

  def main(args: Array[String]) {

    val arr: Array[Array[Int]] = ???

    val sums: Seq[Int] = for {
      i <- 0 until arr.length - 2
      j <- 0 until arr.length - 2
    } yield checkSum(arr, i,     j,     j + 2) + 
            checkSum(arr, i + 1, j + 1, j + 1) +
            checkSum(arr, i + 2, j,     j + 2)

    val res = (sums :+ -2147483648).max

    println(res)
  }

  // as suggested in comment:
  private def checkSum(arr: Array[Array[Int]], row: Int, first: Int, last: Int) = {
    arr(row).slice(first, last+1).sum
  }
}

Answer 2

您可以使用两个foldLeft编写第一个函数：

val res = -2147483648
      (0 until (arr.length-2)).foldLeft(res) { case (acc, i) =>
        (0 until (arr.length-2)).foldLeft(acc) { case (acc2, j) =>
          val sum = checkSum(arr, i, j, j + 2) + checkSum(arr,  i + 1, j + 1, j + 1) + checkSum(arr, i + 2, j, j + 2)
          if(sum > acc2) sum else acc2
        }
        }

您可以使用foldLeft编写秒函数：

(first until (last+1)).foldLeft(0) { case (acc, i) => acc + arr(row)(i)}

正如@TzachZohar所提到的，有一些打字问题！

具有不可变状态的2D数组处理

2 个答案: