这取决于字体。例如，交叉0所占的空间要比普通0大得多。

但如果要猜的话，我会选X或B。

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。

根据您的平台，可能有一种方法从字符串或DrawText()函数中“getWidth”，以某种方式使用width属性。

我会做一个简单的算法时间，利用所需的字体，然后通过alfabet运行，并将其存储在一个小配置或只是计算它在初始化作为一个循环从a到Z并不难。

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';

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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

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;
}