Question

我想以螺旋顺序打印数组。适用于尺寸为3x3,4x4等的阵列。我的代码工作正常，但对于3x5,4x6或5x8大小，输出错误，只返回第一次迭代。

这是我的简单代码：

private function _spiral($rows, $cols, array $array) {
    $offset = 0;
    while($offset < ($rows - 1)){
        for($col = $offset; $col <= $cols - 1; $col++){
            print($array[$offset][$col] . ' ');
        }
        $offset++;
        $cols--;
        for($row = $offset; $row < $rows; $row++){
            print($array[$row][$cols] . ' '); 
        }
        $rows--;
        for($col = $cols - 1; $col >= $offset; $col--){
            print($array[$rows][$col] . ' ');
        }
        for($row = $rows; $row >= $offset; $row--){
            print($array[$row][$offset - 1] . ' ');
        }
    } 
 }

包含3行和4列的示例：

$array = array(
    array(00,01,02,03),
    array(10,11,12,13),
    array(20,21,22,23)
)

此数组的预期结果为0 1 2 3 13 23 22 21 20 10 11 12，但我的函数输出在10之后停止。

4行4列：

$array = array(
    array(00,01,02,03),
    array(10,11,12,13),
    array(20,21,22,23),
    array(30,31,32,33)
)

...它应该返回0 1 2 3 13 23 33 32 31 30 20 10 11 12 22 21，这就是我的代码返回的内容。

但我希望这两种情况都适用于我的代码。如何更正代码以便为第一个和其他情况生成正确的输出？

Answer 1

您的代码存在一些问题：

它不会以同样的方式处理四个遍历方向。这四个方向有四个循环，但在某些方面你有<=作为其他<的循环结束条件，在某些情况下，条件是减去1，而在其他情况下不是。
它没有规定何时所有元素都由第一个或第二个内环打印，因此剩余的循环在某些情况下会打印已打印的元素。
外部循环条件不检查是否仍有需要遍历的列。仅仅测试这样的行是不够的。

虽然您可以尝试修复代码，但我认为最好从头开始，考虑到解决方案应该对所有四个方向都是对称的。这是一个重要的直观反应：点对称性。这将导致更少的代码和更少的错误。

您希望遍历数组中的维（行或列），直到到达数组的边框或已打印的元素。然后你想向右转90°并反复重复完全相同的逻辑。因此，如果您的代码对于这些不同的方向看起来不同，那么事情就不对了。

我将分享两个实现。两者都将使用“当前”单元格的概念，让它以螺旋运动的方式移动。

第一个解决方案使用相同的代码处理沿着一行返回或前进，类似地，它有一段用于向前或向后遍历列的代码。所以这个解决方案有两个内部循环，一个用于沿着一行遍历，另一个用于沿着列遍历。遍历行或列的方向保存在$direction变量中，每次执行外循环时，该变量在1和-1之间翻转：

function _spiral(array $array) {
    // No need to have the number of rows and columns passed as arguments:
    // We can get that information from the array:
    $rows = count($array);
    $cols = count($array[0]);
    // Set "current" cell to be outside array: it moves into it in first inner loop
    $row = 0;
    $col = -1;
    $direction = 1; // Can be 1 for forward and -1 for backward
    while ($rows > 0 and $cols > 0) {
        // Print cells along one row
        for ($step = 0; $step < $cols; $step++) {
            $col += $direction;
            print $array[$row][$col] . ' ';
        }
        // As we have printed a row, we have fewer rows left to print from:
        $rows--;
        // Print cells along one column
        for ($step = 0; $step < $rows; $step++) {
            $row += $direction;
            print $array[$row][$col] . ' ';
        }
        // As we have printed a column, we have fewer columns left to print from:
        $cols--;
        // Now flip the direction between forward and backward
        $direction = -$direction;
    }
}

注意第一个内环和第二个内环之间的完美对称。

在第二种解决方案中，这种对称性的使用更进一步，以便仅用一个替换两个内环。为此，我们必须放弃对行和列使用单独的变量，并使用与维度相关的大小的概念：

function _spiral(array $array) {
    // This version of the function aims to treat rows and columns in the same way,
    // They are just another dimension, but all the logic is exactly the same:
    // $size[] has the number of rows in $size[0] and number of columns in $size[1]
    $size = Array(count($array), count($array[0]));
    // $current[] has the current row in $current[0] and current column in $current[1]
    $current = Array(0, -1);
    // $direction[] has the current row-traversal direction in $direction[0] 
    //    and column-traveral direction in $direction[1]
    $direction = Array(1, 1);
    $dimension = 0; // Which dimension to traverse along, can be 0 for row, 1 for column
    while ($size[$dimension] > 0)   {
        // Switch dimension (row to column, column to row), to traverse along
        $dimension = 1 - $dimension;
        // Print one line along that dimension, in its current direction
        for ($step = 0; $step < $size[$dimension]; $step++) {
            $current[$dimension] += $direction[$dimension];
            print $array[$current[0]][$current[1]] . ' ';
        }
        // As we have printed a line, we have fewer left to print from:
        $size[1 - $dimension]--;
        // Now flip the direction between forward and backward for this dimension:
        $direction[$dimension] = -$direction[$dimension];
    }
}

扩展版

超过一年后的请求：这是一个版本，允许人们选择从角落开始，以及是逆时针而不是顺时针。此函数不会打印结果，而是返回带有螺旋序列的1D数组。这样你就可以自己决定如何处理结果：打印它，或者......无论如何。

function spiral(array $array, $startRight = false, $startBottom = false, 
                              $counterClockWise = false) {
    // This version allows to select which corner to start from, and in which direction.
    //   $size[] has the number of rows in $size[0] and number of columns in $size[1]
    $size = [count($array), count($array[0])];
    // $direction[] has the current row-traversal direction in $direction[0] 
    //    and column-traversal direction in $direction[1]
    $direction = [$startBottom ? -1 : 1, $startRight ? -1 : 1];
    // Which dimension to traverse along: false means row, true means column.
    //   Every one of the optional arguments will flip the first dimension to use:
    $dimension = ($startBottom xor $startRight xor $counterClockWise);
    // $current[] has the current row in $current[0] and current column in $current[1]
    $current = [$startBottom * (count($array)-1), $startRight * (count($array[0])-1)];
    // Go back one step, outside of the grid
    $current[!$dimension] -= $direction[!$dimension];
    while ($size[$dimension] > 0)   {
        // Switch dimension (row to column, column to row), to traverse along
        $dimension = !$dimension;
        // Print one line along that dimension, in its current direction
        for ($step = 0; $step < $size[$dimension]; $step++) {
            $current[$dimension] += $direction[$dimension];
            $result[] = $array[$current[0]][$current[1]]; // store in new array
        }
        // As we have printed a line, we have fewer left to print from:
        $size[!$dimension]--;
        // Now flip the direction between forward and backward for this dimension:
        $direction[$dimension] = -$direction[$dimension];
    }
    return $result; // Return the resulting spiral as a 1D array
}

在eval.in

上查看它

以螺旋顺序打印2D数组值

1 个答案:

扩展版