invoker | 技不如人，甘拜下风
2018-03-17

求解：顺时针输出二维数组

给定一个 n*n 的二维数组矩阵矩阵，以顺时针顺序输出

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举个栗子：
[ 0, 1, 2,
  3, 4, 5,
  6, 7, 8 ]
输出： 0, 1, 2, 5, 8, 7, 6, 3, 4

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fun main(args: Array<String>) {
    val cursor = Cursor(-1, 0)
    val rect = makeRect(3)
    rect.print()

    clockwise(rect, 0, cursor)
}

typealias Matrix = Array<IntArray>

data class Cursor(var x: Int, var y: Int) {
    fun up() = apply { y -= 1 }
    fun down() = apply { y += 1 }
    fun left() = apply { x -= 1 }
    fun right() = apply { x += 1 }
}

fun clockwise(rect: Matrix, level: Int, cursor: Cursor) {
    val rectLength = rect[0].size
    val currentLength = rectLength - 2 * level

    for (i in 0 until currentLength) {
        cursor.right()
        print(rect[cursor])
    }
    for (i in 0 until currentLength - 1) {
        cursor.down()
        print(rect[cursor])
    }
    for (i in 0 until currentLength - 1) {
        cursor.left()
        print(rect[cursor])
    }
    for (i in 0 until currentLength - 2) {
        cursor.up()
        print(rect[cursor])
    }

    if (currentLength < 0) {
        return
    } else {
        clockwise(rect, level + 1, cursor)
    }
}

private fun makeRect(n: Int) = Array(n) { IntArray(n) }.apply {
    for (i in 0 until n * n) {
        this[i / n][i % n] = i
    }
}

private fun Matrix.print() = forEach { println(it.joinToString(",")) }
private operator fun Matrix.get(cursor: Cursor) = this[cursor.y][cursor.x]
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