Question

我有一个代表网格的列表列表。我需要访问特定块的元素。例如：

| 0 | 1 | 0 || 0 | 0 | 0 || 0 | 0 | 8 |
| 0 | 0 | 0 || 2 | 9 | 0 || 5 | 1 | 3 |
| 0 | 0 | 0 || 7 | 0 | 0 || 0 | 0 | 0 |
 -------------------------------------
| 0 | 4 | 0 || 5 | 0 | 0 || 7 | 0 | 2 |
| 0 | 0 | 0 || 4 | 6 | 0 || 9 | 8 | 1 |
| 3 | 0 | 0 || 1 | 0 | 0 || 0 | 0 | 0 |
 -------------------------------------
| 0 | 1 | 0 || 5 | 0 | 0 || 4 | 0 | 2 |
| 0 | 2 | 0 || 0 | 1 | 0 || 0 | 0 | 0 |
| 0 | 9 | 0 || 3 | 0 | 0 || 0 | 0 | 4 |

给定块（0）应返回List（1），给定块（5）应返回List（7,8,9,1,2）

我的尝试：

((grid.grouped(3).toList(block / 3).map(_.grouped(3).toList(block % 3)))flatten)

并且它有效，但我需要找到一种没有所有这些集合的简单方法

Answer 1

grid.drop(3 * (block / 3)).take(3).flatMap(_.drop(3 * (block % 3)).take(3)).filter(_ != 0)

这样效率更高 - 直接“缩放”到您想要的数据;坚持列表（没有转换）。

scala> val grid = List(List(0,1,0,0,0,0,0,0,8),
                       List(0,0,0,2,9,0,5,1,3),
                       List(0,0,0,7,0,0,0,0,0),
                       List(0,4,0,5,0,0,7,0,2),
                       List(0,0,0,4,6,0,9,8,1),
                       List(3,0,0,1,0,0,0,0,0),
                       List(0,1,0,5,0,0,4,0,2),
                       List(0,2,0,0,1,0,0,0,0),
                       List(0,9,0,3,0,0,0,0,2))
grid: List[List[Int]] = List(List(0, 1, 0, 0, 0, 0, 0, 0, 8), List(0, 0, 0, 2, 9, 0, 5, 1, 3), List(0, 0, 0, 7, 0, 0, 0, 0, 0), List(0, 4, 0, 5, 0, 0, 7, 0, 2), List(0, 0, 0, 4, 6, 0, 9, 8, 1), List(3, 0, 0, 1, 0, 0, 0, 0, 0), List(0, 1, 0, 5, 0, 0, 4, 0, 2), List(0, 2, 0, 0, 1, 0, 0, 0, 0), List(0, 9, 0, 3, 0, 0, 0, 0, 2))

scala> def blockElements(block: Int) = grid.drop(3 * (block / 3)).take(3).flatMap(_.drop(3 * (block % 3)).take(3)).filter(_ != 0)
blockElements: (block: Int)List[Int]

scala> blockElements(0)
res1: List[Int] = List(1)

scala> blockElements(5)
res2: List[Int] = List(7, 2, 9, 8, 1)

顺便说一句：第9个子广场有两个4：:)。

访问块的Scala中的列表列表

1 个答案: