我正在尝试做一些基于句子中字符数量的动态规划。英语字母表中哪个字母在屏幕上占像素最多?
当前回答
这取决于字体。我会用你最熟悉的编程语言创建一个小程序，把字母表中的每个字母画成n乘以m的位图。用白色初始化每个像素。然后，在你画完每个字母后，数一数白色像素的数量，并保存这个数字。你找到的最大的数字就是你要找的数字。
编辑:如果你实际上只是对哪个占据了最大的矩形感兴趣(但看起来你真的是在追求它，而不是像素)，你可以使用各种API调用来找到大小，但这取决于你的编程语言。例如，在Java中，您将使用FontMetrics类。
其他回答
我认为字母W是最宽的。
我知道公认的答案是W, W代表胜利。
但是，在本例中，W也表示宽度。案例研究使用了一个简单的宽度测试来检查像素，但它只是宽度，而不是总像素数。作为一个简单的反例，公认的答案假设O和Q占用相同数量的像素，但它们只占用相同数量的空间。
因此，W占的空间最大。但是，这是所有的像素，它被吹捧吗?
让我们来看看一些经验数据。我从下面的B, M和w中创建了imgur图像，然后分析了它们的像素数(见下文)，结果如下:
B: 114像素
M: 150像素
女:157像素
以下是我如何将它们放入画布并分析图像中的原始像素数据。
var imgs = { B : "//i.imgur.com/YOuEPOn.png", M : "//i.imgur.com/Aev3ZKQ.png", W : "//i.imgur.com/xSUwE7w.png" }; window.onload = function(){ for(var key in imgs){(function(img,key){ var Out = document.querySelector("#"+key+"Out"); img.crossOrigin = "Anonymous"; img.src=imgs[key]; img.onload = function() { var canvas = document.querySelector('#'+key); (canvas.width = img.width,canvas.height = img.height); var context = canvas.getContext('2d'); context.drawImage(img, 0, 0); var data = context.getImageData(0, 0, img.width, img.height).data; Out.innerHTML = "Total Pixels: " + data.length/4 + "<br>"; var pixelObject = {}; for(var i = 0; i < data.length; i += 4){ var rgba = "rgba("+data[i]+","+data[i+1]+","+data[i+2]+","+data[i+3]+")"; pixelObject[rgba] = pixelObject[rgba] ? pixelObject[rgba]+1 : 1; } Out.innerHTML += "Total Whitespace: " + pixelObject["rgba(255,255,255,255)"] + "<br>"; Out.innerHTML += "Total Pixels In "+ key +": " + ((data.length/4)-pixelObject["rgba(255,255,255,255)"]) + "<br>"; }; })(new Image(),key)} }; <table> <tr> <td> <canvas id="B" width="100%" height="100%"></canvas> </td> <td id="BOut"> </td> </tr> <tr> <td> <canvas id="M" width="100%" height="100%"></canvas> </td> <td id="MOut"> </td> </tr> <tr> <td> <canvas id="W" width="100%" height="100%"></canvas> </td> <td id="WOut"> </td> </tr> </table>
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W获胜。
当然，这是一个愚蠢的经验实验。哪个字母最宽没有唯一的答案。这取决于字体。所以你必须做一个类似的经验实验来找出你所处环境的答案。但事实是，大多数字体遵循相同的惯例，大写W将是最宽的。
主旨与这些字符宽度的比率形式(W = 100)在这里捕获使用特定的示例字体:
https://gist.github.com/imaurer/d330e68e70180c985b380f25e195b90c
那么程序化的解决方案呢?
var capsIndex = 65; var smallIndex = 97 var div = document.createElement('div'); div.style.float = 'left'; document.body.appendChild(div); var highestWidth = 0; var elem; for(var i = capsIndex; i < capsIndex + 26; i++) { div.innerText = String.fromCharCode(i); var computedWidth = window.getComputedStyle(div, null).getPropertyValue("width"); if(highestWidth < parseFloat(computedWidth)) { highestWidth = parseFloat(computedWidth); elem = String.fromCharCode(i); } } for(var i = smallIndex; i < smallIndex + 26; i++) { div.innerText = String.fromCharCode(i); var computedWidth = window.getComputedStyle(div, null).getPropertyValue("width"); if(highestWidth < parseFloat(computedWidth)) { highestWidth = parseFloat(computedWidth); elem = String.fromCharCode(i); } } div.innerHTML = '<b>' + elem + '</b>' + ' won';
这也取决于字体。我在1或2年前用Processing和Helvetica做过这个，它是ILJTYFVCPAXUZKHSEDORGNBQMW，按增加像素的顺序。这个想法是用你正在看的字体在画布上绘制文本，计算像素，然后用HashMap或Dictionary排序。
当然，这可能与您的使用没有直接关系，因为它计算像素面积而不仅仅是宽度。可能也有点过头了。
void setup() {
size(30,30);
HashMap hm = new HashMap();
fill(255);
PFont font = loadFont("Helvetica-20.vlw");
textFont(font,20);
textAlign(CENTER);
for (int i=65; i<91; i++) {
background(0);
text(char(i),width/2,height-(textDescent()+textAscent())/2);
loadPixels();
int white=0;
for (int k=0; k<pixels.length; k++) {
white+=red(pixels[k]);
}
hm.put(char(i),white);
}
HashMap sorted = getSortedMap(hm);
String asciiString = new String();
for (Iterator<Map.Entry> i = sorted.entrySet().iterator(); i.hasNext();) {
Map.Entry me = (Map.Entry)i.next();
asciiString += me.getKey();
}
println(asciiString); //the string in ascending pixel order
}
public HashMap getSortedMap(HashMap hmap) {
HashMap map = new LinkedHashMap();
List mapKeys = new ArrayList(hmap.keySet());
List mapValues = new ArrayList(hmap.values());
TreeSet sortedSet = new TreeSet(mapValues);
Object[] sortedArray = sortedSet.toArray();
int size = sortedArray.length;
// a) Ascending sort
for (int i=0; i<size; i++) {
map.put(mapKeys.get(mapValues.indexOf(sortedArray[i])), sortedArray[i]);
}
return map;
}
