问题描述
我最近参加了一家公司赞助的编码比赛，有一个我不明白的问题，它问的是什么.
I had recently participated in coding competion sponsored by an company and there was this one question which I did not understood, as to what was it asking.
问题来了:
字符串paypal 是更快、更安全的汇款方式"写成从左上角开始的正方形内的顺时针螺旋图案:(您可能希望以固定字体显示此图案以提高可读性).
The string "paypal is the faster, safer way to send money" is written in a clockwise spiral pattern inside a square starting from the upper left corner: (you may want to display this pattern in a fixed font for better legibility).
P A Y P A L
F E R W A I
A M O N Y S
S D Y E T T
R N E S O H
E T S A F E
然后逐行读取:PAYPALFERWAIAMONYSSDYETTRNESOHETSAFE
Then read line after line: PAYPALFERWAIAMONYSSDYETTRNESOHETSAFE
编写接受字符串的代码，计算最小平方包含它并返回转换后的字符串:
Write the code that will take a string, calculate the minimum square that will contain it and return the converted string:
字符串转换(字符串文本)；
String convert(String text);
示例:
convert("paypalisthefastersaferwaytosendmoney")
should return "paypalferwaiamonyssdyettrnesohetsafe"
您了解我们如何解决这个问题吗?
Do you understand as to how we can approach this problem?
推荐答案
我认为问题，如所写，应解释如下:
I believe that the question, as written, is meant to be interpreted as follows:
给定一个字符串，并希望将该字符串作为螺旋线写入方形网格.编写一个函数，找到可以容纳字符串的最小方格，将字符串按顺时针方向围绕网格旋转，将字符串写入网格，最后将行连接在一起.
You are given a string and want to write that string as a spiral into a square grid. Write a function that finds the smallest square that can hold the string, writes the string into the grid by spiraling its characters around the grid clockwise, and finally concatenates the rows together.
例如，字符串In a spiral"看起来像这样:
As an example, the string "In a spiral" would look like this:
I N A
In a spiral -> A L S -> INAALSRIP
R I P
要查看网格的来源，请注意，如果您是这样阅读的:
To see where the grid comes from, note that if you read it like this:
I -> N -> A
|
v
A -> L S
^ |
| v
R <- I <- P
您会返回初始文本，如果将行INA"、ALS"和RIP"粘贴到单个字符串中，则会返回INAALSRIP".
You get back the initial text, and if you glue the rows "INA", "ALS," and "RIP" into a single string you get back "INAALSRIP."
让我们分别考虑每个问题.首先，要查看可以容纳文本的矩形的最小尺寸，您实际上是在寻找至少与文本长度一样大的最小完美正方形.要找到这个，您可以取字符串长度的平方根并将其四舍五入到最接近的整数.这为您提供了您想要的尺寸.但是，在您这样做之前，您需要从字符串中删除所有标点符号和空格字符(可能还有数字，具体取决于应用程序).您可以通过遍历字符串并将确实是字母的字符复制到新缓冲区中来做到这一点.在接下来的内容中，我假设您已经完成了这项工作.
Let's consider each problem separately. First, to see the smallest size of a rectangle that can hold the text, you're essentially looking for the smallest perfect square at least as large as the length of your text. To find this, you could take the square root of the length of your string and round it up to the nearest integer. That gives you the dimension you'd like. However, before you can do that, you need to strip out all of the punctuation and space characters from the string (and perhaps the numbers as well, depending on the application). You could do this by walking across the string and copying over the characters that are indeed alphabetic into a new buffer. In what follows, I'll assume that you've done this.
至于如何实际填充网格，有一个非常棒的方法可以做到这一点.直觉如下.当您从 n x n 网格开始时，您唯一的边界是网格的墙壁.每次你穿过网格放下字母并撞到墙上时，你只是从矩阵中刮掉了一行或一列.因此，您的算法可以通过跟踪第一个和最后一个合法列以及第一个和最后一个合法行来工作.然后，您从左到右走过顶行书写字符.完成后，然后增加第一个合法行，因为你不能再放任何东西了.然后，沿着右侧向下走，直到到达底部.完成后，您也可以将最后一列排除在外.例如，回顾我们的螺旋形"示例，我们从一个空的 3x3 网格开始:
As for how to actually fill in the grid, there's a really great way to do this. The intuition is as follows. When you start off in an n x n grid, your only boundaries are the walls of the grid. Every time you march across the grid dropping letters and hit a wall, you've just shaved off a row or a column from the matrix. Consequently, your algorithm could work by keeping track of the first and last legal column and the first and last legal row. You then walk across the top row from left to right writing characters. When you're done, you then increment the first legal row, since you can't put anything there any more. Then, walk down the right side until you hit the bottom. Once you're done, you would then block off the last column from consideration as well. For example, to look back at our "In a spiral" example, we start off with an empty 3x3 grid:
. . .
. . .
. . .
在顶部写下前三个字符后，剩下的是:
After we write the first three characters across the top, we're left with this:
I N A
. . .
. . .
现在，我们需要将字符串的其余部分写入空白区域，从右上角开始向下移动.因为我们永远无法写回第一行，所以一种思考方式是考虑解决在较小的空间中以螺旋形式书写其余字符的问题
Now, we need to write the rest of the string into the blank space, starting from the upper-right square and moving down. Because we can't ever write back into the top row, one way of thinking about this is to think about solving the problem of writing the rest of the characters in a spiral in the smaller space
. . .
. . .
从左上角开始向下移动.
Starting from the upper-left corner and moving downward.
要真正将其实现为一种算法，我们需要在每个点跟踪一些事情.首先，我们需要在更新它们时存储世界的边界.我们还需要存储我们当前的写入位置，以及我们面临的方向.在伪代码中，这表示如下:
To actually realize this as an algorithm, we need to keep track of a few things at each point. First, we need to store the bounds of the world as we're updating them. We also need to store our current write location, along with what direction we're facing. In pseudocode, this is represented as follows:
firstRow = 0, lastRow = N - 1 // Bounds of the grid
firstCol = 0, lastCol = N - 1
dRow = 0 // Amount to move in the Y direction
dCol = 1 // Amount to move in the X direction
row = 0 // Current position
col = 0
for each character ch in the string:
Write character ch to position (row, col).
// See if we're blocked and need to turn.
If (row + dRow, col + dCol) is not contained in the rectangle [firstRow, lastRow] x [firstCol, lastCol]:
// Based on which way we are currently facing, adjust the bounds of the world.
If moving left, increment firstRow
If moving down, decrement lastCol
If moving right, decrement lastRow
If moving up, increment firstCol
Rotate 90 degrees
// Finally, move forward a step.
row += dRow
col += dCol
您可以使用线性代数中的一个技巧来实现 90 度转弯:将向量向左旋转 90 度，将其乘以 旋转矩阵
You can implement the ninety-degree turn using a trick from linear algebra: to rotate a vector 90 degrees to the left, you multiply it by the rotation matrix
| 0 1 |
| -1 0 |
所以你的新 dy 和 dx 由
So your new dy and dx are given by
|dCol'| = | 0 1 | dCol = |-dRow|
|dRow'| | -1 0 | dRow | dCol|
所以你可以通过计算左转
So you can turn left by computing
temp = dCol;
dCol = -dRow;
dRow = temp;
另外，如果您知道数值为零的字符永远不会出现在字符串中，您可以使用 Java 初始化所有数组以在任何地方都保持零的事实.然后，您可以将 0 视为哨兵，意思是继续前进是安全的".该版本的(伪)代码如下所示:
Alternatively, if you know for a fact that the character with numeric value zero never appears in the string, you can use the fact that Java initializes all arrays to hold zeros everywhere. You can then treat 0 as a sentinel meaning "it's safe to keep moving forward." That version of the (pseudo)code would look like this:
dRow = 0 // Amount to move in the X direction
dCol = 1 // Amount to move in the Y direction
row = 0 // Current position
col = 0
for each character ch in the string:
Write character ch to position (row, col).
If (row + dRow, col + dCol) is not contained in the rectangle [0, 0] x [n-1, n-1]
-or-
The character at [row + dRow, col + dCol] is not zero:
Rotate 90 degrees
// Move forward a step
row += dRow
col += dCol
最后，将字符串写入螺旋后，您可以通过一次一行地遍历行并将找到的所有字符连接在一起，将螺旋文本转换回字符串.
Finally, once you've written the string into the spiral, you can convert that spiraled text back into a string by walking across the rows one at a time and concatenating together all of the characters that you find.
编辑:正如@Voo 指出的那样，您可以通过根本不实际创建多维数组而是将多维数组编码为一维数组来简化该算法的最后一步.这是一个常见的(而且很聪明！)技巧.例如，假设我们有一个这样的网格:
EDIT: As @Voo points out, you can simplify the last step of this algorithm by not actually creating a multidimensional array at all and instead encoding the multidimensional array as a single-dimensional array. This is a common (and clever!) trick. Suppose, for example, that we have a grid like this:
0 1 2
3 4 5
6 7 8
那么我们可以用一维数组来表示这个
Then we can represent this using a one-dimensional array as
0 1 2 3 4 5 6 7 8
这个想法是，给定 N x N 网格中的 (row, col) 对，我们可以通过查看位置 row * N + col 将该坐标转换为线性化数组中的相应位置.直观地说，这表示你在 y 方向上的每一步都相当于跳过一行中的所有 N 个元素，并且每一个水平步骤只是在线性化表示中水平移动一个步骤.
The idea is that given an (row, col) pair in an N x N grid, we can convert that coordinate to a corresponding location in the linearized array by looking at position row * N + col. Intuitively, this says that every step in the y direction that you take is equivalent to skipping all N elements of one row, and each horizontal step just moves one step horizontally in the linearized representation.
希望这会有所帮助！
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