找出有限区域内字符的最大像素大小
Finding out the maximum pixel size of characters in a limited area
我正在尝试实现一项功能，以显示适合该区域的最大可能字符像素数。
示例 1：
想象一下尺寸为 100x20 像素的 LED 显示屏。需要显示的字符串是：volutpa commodo diam
手工完成所有操作后，我能够确定没有 splitting/dividing 单词的字符的最大可能像素大小为 8x8 像素。每个字符必须具有相同的像素大小。
如果字符像素大小为8x8 px，则上述字符串将成功显示在LED显示屏上。预期输出：8
示例 2：
想象一下尺寸为 100x20 像素的 LED 显示屏。需要显示的字符串是：dio volutpa diam
如果字符像素大小为9x9 px，则上述字符串将成功显示在LED显示屏上。预期输出：9
示例 3：
想象一下尺寸为 55x25 像素的 LED 显示屏。需要显示的字符串是：dio tuo diam
如果字符像素大小为8x8 px，则上述字符串将成功显示在LED显示屏上。预期输出：8
示例 4：
想象一下尺寸为 20x6 像素的 LED 显示屏。需要显示的一个字符串是：tuo volutpa
如果字符像素大小为2x2 px，则上述字符串将成功显示在LED显示屏上。预期输出：2
注意：字符串中的每个字符（包括space）必须具有相同的像素大小（例如 6x6 像素）以填充整个区域（LED 显示屏）并且字符串中的单词不能是 split/divided.
在此 example 1 中，您可以看到我正在努力实现的 "hand made" 图像。整个布局代表区域大小（100x20 像素），而黑框代表最大可能的字符像素大小（8x8 像素）。如果我使用（9x9 px 或更多）作为字符大小，文本将不适合，如果我使用（7x7 px 或更小），则会有太多空白 space.
在此 example 2 中，您可以看到我正在努力实现的 "hand made" 图像。整个布局代表区域大小（100x20 像素），而黑框代表最大可能的字符像素大小（9x9 像素）。如果我使用（10x10 像素或更多）作为字符大小，文本将不适合，如果我使用（8x8 像素或更小），则会有太多空白 space。如果该行即将结束，我们应该忽略单词末尾的 space。
预计：
第 1 行：dio_volutpa
第 2 行：直径
实际:
第 1 行：dio_volutpa_
第 2 行：直径
在此 example 3 中，您可以看到我正在努力实现的 "hand made" 图像。整个布局代表区域大小（55x25 像素），而黑框代表最大可能的字符像素大小（8x8 像素）。如果我使用（9x9 像素或更多）作为字符大小，文本将不适合，如果我使用（7x7 像素或更小），则会有太多空白 space。那里的红框告诉你下一个词没有 space，因此它需要换行或者如果下面没有行 space 则减小它的像素大小。
在 C# 中实现它的最佳方法是什么？
static void Main(string[] args)
{
String sentence = "volutpa commodo diam";
int width = 100; // 100 px
int height = 20; // 20 px
int pixelSize = height / 2; // Starting pixel size (10)
int maxLines = height / pixelSize; // Max. available lines in the area
int currentLine = 1; // Starting line
string[] words = sentence.Split(' '); // Splitting words from the sentence. words[1] = volutpa, words[2] = commodo, words[3] = diam
// Length of the words
for (int i = 0; i < words.Length; i++)
{
Console.WriteLine("Word " + i + " length: " + words[i].Length * pixelSize);
}
Console.ReadLine();
}
我现在已经做了这个，但不知道如何继续。我会很感激一些提示和想法。
这是我根据您更新后的问题得出的最终解决方案。
请阅读代码中的注释。
UPDATE
inside fixtoFitMatrix() 方法 space 将仅在 2 个单词之间被考虑。
更新 2
请阅读代码中的注释1、注释2和注释3。
namespace ConsoleApplication29
{
class Program
{
private static void Main(string[] args)
{
Check("volutpa commodo diam", 100, 20);
Check("dio volutpa diam", 100, 20);
Check("dio tuo diam", 55, 25);
Check("tuo volutpa", 20, 6);
Check("volutpa", 23, 2);
Check("diam tu diam op comodo", 10, 20);
Check("volutpas di diam e commodo", 65, 65);
Check("diamo volutpasoni onedo commodoso", 10, 40);
Console.ReadLine();
}
private static void Check(string txt,int w,int h)
{
String sentence = txt;
int width = w;
int height = h;
string[] words = sentence.Split(' ');
int maxLines;
int pixelSize = GetPixelSize(words, height, width,out maxLines);
Console.WriteLine(sentence);
// length of the words
for (int i = 0; i < words.Length; i++)
{
Console.WriteLine("Word " + i + " length: " + words[i].Length * pixelSize );
}
Console.WriteLine("pixel size will be " + pixelSize + " lines: " + maxLines );
Console.WriteLine("--------------------------------------------");
}
private static int GetPixelSize(string[] words, int height, int width,out int maxlines)
{
int lines = 1;
int basepx = GetBasePixelSize(words, height, width,1,ref lines);
// fix if not each line is fitted to width or the height of the matrix
int finalpx = fixtoFitMatrix(basepx,words,width,height,lines);
maxlines = lines;
return finalpx;
}
private static int fixtoFitMatrix(int basepx, string[] words,int width,int height,int lines)
{
List<LedLine> ldlines = new List<LedLine>();
int currentline = 0;
// set text for each line
LedLine l = new LedLine() { LineNumber = currentline, LineText = words[0]};
ldlines.Add(l);
for (int i = 1;i< words.Length;i++)
{
var currentstring = words[i];
if (ldlines[currentline].Len * basepx + basepx + currentstring.Length * basepx <= width)
{
// add to current line
ldlines[currentline].LineText = ldlines[currentline].LineText + currentstring;
}
else
{
// it is a new line
currentline++;
LedLine newl = new LedLine() { LineNumber = currentline, LineText = currentstring };
ldlines.Add(newl);
}
}
// px size is also a function of the width
// for example:
// on surface of 100X20 volutpa commodo diam will be 8px and dio volutpa diam will be 9px
// that happens because commodo diam is the length of 12 and 12*9=108 and the width is only 100 so we will need to fix
//fix will be chcked if the number of lines requeired is not the same in ldlines.count that means that we will need to "shrink" px by 1 every recursive call
if (ldlines.Count > lines)
{
return fixtoFitMatrix(basepx - 1, words, width,height, lines);
}
// comment 1: we need to check the height also
else if (lines * basepx > height)
{
return fixtoFitMatrix(basepx - 1, words, width, height, lines);
}
return basepx;
}
private static int GetBasePixelSize(string[] words, int height,int width,int linesneeded,ref int finalnumberoflines,int specialH = 0)
{
int h = (int)Math.Ceiling(((double)height / (double)linesneeded));
if (specialH > 0)
h = specialH;
var lines = linesneeded;
var accumulate = 0;
for (int i = 0;i< words.Length;i++)
{
accumulate += words[i].Length * h;
// add space if it is not the last word
if (i < words.Length - 1)
{
accumulate += h;
}
// more line needed - call recursivaly until we get the number of lines needed
if (accumulate / lines > width)
{
return GetBasePixelSize(words, height, width, linesneeded+1,ref finalnumberoflines,specialH);
}
}
finalnumberoflines = lines;
// comment 2 - special case - if there is a single word that is larger than all px (int h) we will decrease h, for example on matrix of 10*20 the word 'comodo'
// can be displayed only when h=1 because if its 2 then 2* 6 = 12
// we will strat all over again with the new size
if (words.Any(x => x.Length * h > width))
{
string specialword = words.Where(x => x.Length * h > width).FirstOrDefault();
// decrease h by the equation X = height/word.length, if the result is under 1 its not possible to do that
h = (int)Math.Floor((double)width/ (double)specialword.Length);
// comment 3 - if h = 0 exit the method and return 0 - it is not possible with the given conditions
// (for example matrix of 10X40 with the sentence 'diamo volutpasoni onedo commodoso' is not possible)
if (h == 0)
return 0;
finalnumberoflines = 1;
return GetBasePixelSize(words, height, width, 1, ref finalnumberoflines,h);
}
return h;
}
}
public class LedLine
{
public int LineNumber { get; set; }
public string LineText { get; set; }
public int Len
{
get
{
return LineText.Length;
}
}
}
}
输出：
volutpa commodo diam
Word 0 length: 56
Word 1 length:
56
Word 2 length: 32
pixel size will be 8 lines: 2
--------------------------------------------
dio volutpa diam
Word 0 length: 27
Word 1 length: 63
Word 2 length:
36
pixel size will be 9 lines: 2
--------------------------------------------
dio tuo diam
Word 0 length: 24
Word 1 length: 24
Word 2 length: 32
pixel size will be 8 lines: 3
--------------------------------------------
tuo volutpa
Word 0 length: 6
Word 1 length: 14
pixel size will be 2
lines: 2
--------------------------------------------
volutpa
Word 0 length: 14
pixel size will be 2 lines: 1
--------------------------------------------
diam tu diam op comodo
Word 0 length: 4
Word 1 length: 2
Word 2 length:
4
Word 3 length: 2
Word 4 length: 6
pixel size will be
1 lines: 3
--------------------------------------------
volutpas di diam e commodo
Word 0 length: 64
Word 1 length: 16
Word 2
length: 32
Word 3 length: 8
Word 4 length: 56
pixel
size will be 8 lines: 4
--------------------------------------------
diamo volutpasoni onedo commodoso
Word 0 length: 0
Word 1 length: 0
Word
2 length: 0
Word 3 length: 0
pixel size will be 0 lines:
7
--------------------------------------------
我正在尝试实现一项功能，以显示适合该区域的最大可能字符像素数。
示例 1： 想象一下尺寸为 100x20 像素的 LED 显示屏。需要显示的字符串是：volutpa commodo diam
手工完成所有操作后，我能够确定没有 splitting/dividing 单词的字符的最大可能像素大小为 8x8 像素。每个字符必须具有相同的像素大小。
如果字符像素大小为8x8 px，则上述字符串将成功显示在LED显示屏上。预期输出：8
示例 2： 想象一下尺寸为 100x20 像素的 LED 显示屏。需要显示的字符串是：dio volutpa diam
如果字符像素大小为9x9 px，则上述字符串将成功显示在LED显示屏上。预期输出：9
示例 3： 想象一下尺寸为 55x25 像素的 LED 显示屏。需要显示的字符串是：dio tuo diam
如果字符像素大小为8x8 px，则上述字符串将成功显示在LED显示屏上。预期输出：8
示例 4： 想象一下尺寸为 20x6 像素的 LED 显示屏。需要显示的一个字符串是：tuo volutpa
如果字符像素大小为2x2 px，则上述字符串将成功显示在LED显示屏上。预期输出：2
注意：字符串中的每个字符（包括space）必须具有相同的像素大小（例如 6x6 像素）以填充整个区域（LED 显示屏）并且字符串中的单词不能是 split/divided.
在此 example 1 中，您可以看到我正在努力实现的 "hand made" 图像。整个布局代表区域大小（100x20 像素），而黑框代表最大可能的字符像素大小（8x8 像素）。如果我使用（9x9 px 或更多）作为字符大小，文本将不适合，如果我使用（7x7 px 或更小），则会有太多空白 space.
在此 example 2 中，您可以看到我正在努力实现的 "hand made" 图像。整个布局代表区域大小（100x20 像素），而黑框代表最大可能的字符像素大小（9x9 像素）。如果我使用（10x10 像素或更多）作为字符大小，文本将不适合，如果我使用（8x8 像素或更小），则会有太多空白 space。如果该行即将结束，我们应该忽略单词末尾的 space。
预计：
第 1 行：dio_volutpa
第 2 行：直径
实际:
第 1 行：dio_volutpa_
第 2 行：直径
在此 example 3 中，您可以看到我正在努力实现的 "hand made" 图像。整个布局代表区域大小（55x25 像素），而黑框代表最大可能的字符像素大小（8x8 像素）。如果我使用（9x9 像素或更多）作为字符大小，文本将不适合，如果我使用（7x7 像素或更小），则会有太多空白 space。那里的红框告诉你下一个词没有 space，因此它需要换行或者如果下面没有行 space 则减小它的像素大小。
在 C# 中实现它的最佳方法是什么？
static void Main(string[] args)
{
String sentence = "volutpa commodo diam";
int width = 100; // 100 px
int height = 20; // 20 px
int pixelSize = height / 2; // Starting pixel size (10)
int maxLines = height / pixelSize; // Max. available lines in the area
int currentLine = 1; // Starting line
string[] words = sentence.Split(' '); // Splitting words from the sentence. words[1] = volutpa, words[2] = commodo, words[3] = diam
// Length of the words
for (int i = 0; i < words.Length; i++)
{
Console.WriteLine("Word " + i + " length: " + words[i].Length * pixelSize);
}
Console.ReadLine();
}
我现在已经做了这个，但不知道如何继续。我会很感激一些提示和想法。
这是我根据您更新后的问题得出的最终解决方案。
请阅读代码中的注释。
UPDATE
inside fixtoFitMatrix() 方法 space 将仅在 2 个单词之间被考虑。
更新 2
请阅读代码中的注释1、注释2和注释3。
namespace ConsoleApplication29
{
class Program
{
private static void Main(string[] args)
{
Check("volutpa commodo diam", 100, 20);
Check("dio volutpa diam", 100, 20);
Check("dio tuo diam", 55, 25);
Check("tuo volutpa", 20, 6);
Check("volutpa", 23, 2);
Check("diam tu diam op comodo", 10, 20);
Check("volutpas di diam e commodo", 65, 65);
Check("diamo volutpasoni onedo commodoso", 10, 40);
Console.ReadLine();
}
private static void Check(string txt,int w,int h)
{
String sentence = txt;
int width = w;
int height = h;
string[] words = sentence.Split(' ');
int maxLines;
int pixelSize = GetPixelSize(words, height, width,out maxLines);
Console.WriteLine(sentence);
// length of the words
for (int i = 0; i < words.Length; i++)
{
Console.WriteLine("Word " + i + " length: " + words[i].Length * pixelSize );
}
Console.WriteLine("pixel size will be " + pixelSize + " lines: " + maxLines );
Console.WriteLine("--------------------------------------------");
}
private static int GetPixelSize(string[] words, int height, int width,out int maxlines)
{
int lines = 1;
int basepx = GetBasePixelSize(words, height, width,1,ref lines);
// fix if not each line is fitted to width or the height of the matrix
int finalpx = fixtoFitMatrix(basepx,words,width,height,lines);
maxlines = lines;
return finalpx;
}
private static int fixtoFitMatrix(int basepx, string[] words,int width,int height,int lines)
{
List<LedLine> ldlines = new List<LedLine>();
int currentline = 0;
// set text for each line
LedLine l = new LedLine() { LineNumber = currentline, LineText = words[0]};
ldlines.Add(l);
for (int i = 1;i< words.Length;i++)
{
var currentstring = words[i];
if (ldlines[currentline].Len * basepx + basepx + currentstring.Length * basepx <= width)
{
// add to current line
ldlines[currentline].LineText = ldlines[currentline].LineText + currentstring;
}
else
{
// it is a new line
currentline++;
LedLine newl = new LedLine() { LineNumber = currentline, LineText = currentstring };
ldlines.Add(newl);
}
}
// px size is also a function of the width
// for example:
// on surface of 100X20 volutpa commodo diam will be 8px and dio volutpa diam will be 9px
// that happens because commodo diam is the length of 12 and 12*9=108 and the width is only 100 so we will need to fix
//fix will be chcked if the number of lines requeired is not the same in ldlines.count that means that we will need to "shrink" px by 1 every recursive call
if (ldlines.Count > lines)
{
return fixtoFitMatrix(basepx - 1, words, width,height, lines);
}
// comment 1: we need to check the height also
else if (lines * basepx > height)
{
return fixtoFitMatrix(basepx - 1, words, width, height, lines);
}
return basepx;
}
private static int GetBasePixelSize(string[] words, int height,int width,int linesneeded,ref int finalnumberoflines,int specialH = 0)
{
int h = (int)Math.Ceiling(((double)height / (double)linesneeded));
if (specialH > 0)
h = specialH;
var lines = linesneeded;
var accumulate = 0;
for (int i = 0;i< words.Length;i++)
{
accumulate += words[i].Length * h;
// add space if it is not the last word
if (i < words.Length - 1)
{
accumulate += h;
}
// more line needed - call recursivaly until we get the number of lines needed
if (accumulate / lines > width)
{
return GetBasePixelSize(words, height, width, linesneeded+1,ref finalnumberoflines,specialH);
}
}
finalnumberoflines = lines;
// comment 2 - special case - if there is a single word that is larger than all px (int h) we will decrease h, for example on matrix of 10*20 the word 'comodo'
// can be displayed only when h=1 because if its 2 then 2* 6 = 12
// we will strat all over again with the new size
if (words.Any(x => x.Length * h > width))
{
string specialword = words.Where(x => x.Length * h > width).FirstOrDefault();
// decrease h by the equation X = height/word.length, if the result is under 1 its not possible to do that
h = (int)Math.Floor((double)width/ (double)specialword.Length);
// comment 3 - if h = 0 exit the method and return 0 - it is not possible with the given conditions
// (for example matrix of 10X40 with the sentence 'diamo volutpasoni onedo commodoso' is not possible)
if (h == 0)
return 0;
finalnumberoflines = 1;
return GetBasePixelSize(words, height, width, 1, ref finalnumberoflines,h);
}
return h;
}
}
public class LedLine
{
public int LineNumber { get; set; }
public string LineText { get; set; }
public int Len
{
get
{
return LineText.Length;
}
}
}
}
输出：
volutpa commodo diam
Word 0 length: 56
Word 1 length: 56
Word 2 length: 32
pixel size will be 8 lines: 2
--------------------------------------------
dio volutpa diam
Word 0 length: 27
Word 1 length: 63
Word 2 length: 36
pixel size will be 9 lines: 2
--------------------------------------------
dio tuo diam
Word 0 length: 24
Word 1 length: 24
Word 2 length: 32
pixel size will be 8 lines: 3
--------------------------------------------
tuo volutpa
Word 0 length: 6
Word 1 length: 14
pixel size will be 2 lines: 2
--------------------------------------------
volutpa
Word 0 length: 14
pixel size will be 2 lines: 1
--------------------------------------------
diam tu diam op comodo
Word 0 length: 4
Word 1 length: 2
Word 2 length: 4
Word 3 length: 2
Word 4 length: 6
pixel size will be 1 lines: 3
--------------------------------------------
volutpas di diam e commodo
Word 0 length: 64
Word 1 length: 16
Word 2 length: 32
Word 3 length: 8
Word 4 length: 56
pixel size will be 8 lines: 4
--------------------------------------------
diamo volutpasoni onedo commodoso
Word 0 length: 0
Word 1 length: 0
Word 2 length: 0
Word 3 length: 0
pixel size will be 0 lines: 7
--------------------------------------------