216 lines
9.9 KiB
Markdown
216 lines
9.9 KiB
Markdown
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---
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title: Roman Numeral Converter
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localeTitle: 罗马数字转换器
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---
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![:triangular_flag_on_post:](https://forum.freecodecamp.com/images/emoji/emoji_one/triangular_flag_on_post.png?v=3 ":triangular_flag_on_post:")如果卡住,请记得使用**`Read-Search-Ask`** 。尝试配对程序![:busts_in_silhouette:](https://forum.freecodecamp.com/images/emoji/emoji_one/busts_in_silhouette.png?v=3 ":busts_in_silhouette:")并编写自己的代码![:pencil:](https://forum.freecodecamp.com/images/emoji/emoji_one/pencil.png?v=3 ":铅笔:")
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### ![:checkered_flag:](https://forum.freecodecamp.com/images/emoji/emoji_one/checkered_flag.png?v=3 ":checkered_flag:")问题说明:
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您将创建一个将整数转换为罗马数字的程序。
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#### 相关链接
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* [罗马数字](http://www.mathsisfun.com/roman-numerals.html)
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* [方法Array.splice()](http://forum.freecodecamp.com/t/javascript-array-prototype-splice/14307)
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* [Array.indexOf()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf)
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* [Array.join()](http://forum.freecodecamp.com/t/javascript-array-prototype-join/14292)
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## ![:speech_balloon:](https://forum.freecodecamp.com/images/emoji/emoji_one/speech_balloon.png?v=3 ":speech_balloon:")提示:1
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创建两个数组,一个使用罗马数字,一个使用十进制等效的新表单将非常有用。
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> _现在尝试解决问题_
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## ![:speech_balloon:](https://forum.freecodecamp.com/images/emoji/emoji_one/speech_balloon.png?v=3 ":speech_balloon:")提示:2
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如果将数字添加到引入新字母之前的数组中,例如4,9和40的值,它将为您节省大量代码。
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> _现在尝试解决问题_
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## ![:speech_balloon:](https://forum.freecodecamp.com/images/emoji/emoji_one/speech_balloon.png?v=3 ":speech_balloon:")提示:3
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您不能将三个以上的连续罗马数字放在一起。
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> _现在尝试解决问题_
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## 扰流警报!
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![警告牌](//discourse-user-assets.s3.amazonaws.com/original/2X/2/2d6c412a50797771301e7ceabd554cef4edcd74d.gif)
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**提前解决!**
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## ![:beginner:](https://forum.freecodecamp.com/images/emoji/emoji_one/beginner.png?v=3 ":初学者:")基本代码解决方案
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```
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var convertToRoman = function(num) {
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var decimalValue = [ 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 ];
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var romanNumeral = [ 'M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I' ];
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var romanized = '';
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for (var index = 0; index < decimalValue.length; index++) {
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while (decimalValue[index] <= num) {
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romanized += romanNumeral[index];
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num -= decimalValue[index];
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}
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}
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return romanized;
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}
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// test here
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convertToRoman(36);
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```
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![:rocket:](https://forum.freecodecamp.com/images/emoji/emoji_one/rocket.png?v=3 ":火箭:") [运行代码](https://repl.it/CLmf/0)
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### 代码说明:
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* 我们首先创建两个具有匹配索引的默认转换的数组。这些被称为`decimalValue`和`romanNumeral` 。我们还创建了一个`romanized`的空字符串变量,它将包含最终的罗马数字。
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* 使用for循环,我们循环遍历`decimalValue`数组的`decimalValue` 。我们继续循环,直到当前`index`的值适合`num` 。
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* 接下来,我们添加罗马数字,并将`num`十进制等值。
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* 最后,我们返回`romanized`的值。
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#### 相关链接
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* [对于循环](http://forum.freecodecamp.com/t/javascript-for-loop/14666)
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* 循环时
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## ![:sunflower:](https://forum.freecodecamp.com/images/emoji/emoji_one/sunflower.png?v=3 ":向日葵:")中级代码解决方案:
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```
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function convertToRoman(num) {
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var romans = ["I", "V", "X", "L", "C", "D", "M"],
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ints = [],
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romanNumber = [],
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numeral = "";
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while (num) {
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ints.push(num % 10);
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num = Math.floor(num/10);
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}
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for (i=0; i<ints.length; i++){
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units(ints[i]);
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}
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function units(){
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numeral = romans[i*2];
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switch(ints[i]) {
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case 1:
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romanNumber.push(numeral);
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break;
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case 2:
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romanNumber.push(numeral.concat(numeral));
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break;
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case 3:
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romanNumber.push(numeral.concat(numeral).concat(numeral));
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break;
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case 4:
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romanNumber.push(numeral.concat(romans[(i*2)+1]));
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break;
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case 5:
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romanNumber.push(romans[(i*2)+1]);
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break;
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case 6:
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romanNumber.push(romans[(i*2)+1].concat(numeral));
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break;
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case 7:
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romanNumber.push(romans[(i*2)+1].concat(numeral).concat(numeral));
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break;
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case 8:
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romanNumber.push(romans[(i*2)+1].concat(numeral).concat(numeral).concat(numeral));
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break;
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case 9:
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romanNumber.push(romans[i*2].concat(romans[(i*2)+2]));
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}
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}
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return romanNumber.reverse().join("").toString();
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}
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// test here
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convertToRoman(97);
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```
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![:rocket:](https://forum.freecodecamp.com/images/emoji/emoji_one/rocket.png?v=3 ":火箭:") [运行代码](https://repl.it/C1YV)
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### 代码说明:
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* 创建一个罗马数字( `romans` )数组。
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* 使用for循环在数字中创建数字( `ints` )数组。
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* 循环遍历数字数组(基数为10),并且如您所做,将罗马数字(基数5)索引增加2( `numeral = romans[i*2]`罗曼`numeral = romans[i*2]` )。
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* 在循环内,使用Switch Case将正确的罗马数字(向后)推送到该数组。
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* 反转Roman Numerals数组并将其转换为字符串。
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#### 相关链接
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* [对于循环](http://forum.freecodecamp.com/t/javascript-for-loop/14666)
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* 循环时
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* [数学](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math)
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## ![:sunflower:](https://forum.freecodecamp.com/images/emoji/emoji_one/sunflower.png?v=3 ":向日葵:")中级代码解决方案:
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```
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function convertToRoman(num) {
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var romans =
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// 10^i 10^i*5
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["I", "V"], // 10^0
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["X", "L"], // 10^1
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["C", "D"], // 10^2
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["M"] // 10^3
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],
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digits = num.toString()
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.split('')
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.reverse()
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.map(function(item, index) {
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return parseInt(item);
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}),
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numeral = "";
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// Loop through each digit, starting with the ones place
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for (var i=0; i<digits.length; i++) {
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// Make a Roman numeral that ignores 5-multiples and shortening rules
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numeral = romans[i][0].repeat(digits[i]) + numeral;
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// Check for a Roman numeral 5-multiple version
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if (romans[i][1]) {
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numeral = numeral
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// Change occurrences of 5 * 10^i to the corresponding 5-multiple Roman numeral
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.replace(romans[i][0].repeat(5), romans[i][1])
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// Shorten occurrences of 9 * 10^i
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.replace(romans[i][1] + romans[i][0].repeat(4), romans[i][0] + romans[i+1][0])
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// Shorten occurrences of 4 * 10^i
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.replace(romans[i][0].repeat(4), romans[i][0] + romans[i][1]);
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}
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}
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return numeral;
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}
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// test here
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convertToRoman(36);
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```
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![:rocket:](https://forum.freecodecamp.com/images/emoji/emoji_one/rocket.png?v=3 ":火箭:") [运行代码](https://repl.it/C1YV)
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### 代码说明:
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* 使用数组( `romans` )创建一个矩阵,其中包含给定功率为10的罗马数字,如果可用,则为该功率的罗马数字为10倍5。
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* 将输入数字( `num` )转换为反转的数字( `digits` )数组,以便我们可以循环显示从那些位置开始向上的数字。
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* 循环通过每个数字,从个位,以及通过将每个高功率罗马数字到开始创建一个罗马数字串`numeral`串等于多次`digit` 。这个初始字符串忽略了10次幂的罗马数字,也忽略了缩短规则。
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* 如果10的相关幂具有5个多重罗马数字,则用`numeral`代替5-in-row次数与相关的5-multiple罗马数字(即V,L或D)并缩短9 \*的出现次数10 ^ i(例如,VIIII至VIX)和4 \* 10 ^ i(例如,XXXX至XL)。订单在这里很重要!
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* 最后,返回`numeral` 。
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#### 相关链接
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* [对于循环](http://forum.freecodecamp.com/t/javascript-for-loop/14666)
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* [。分裂()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/split)
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* [。相反()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reverse)
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* [。地图()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map)
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* [的ToString()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/toString)
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* [parseInt函数()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/parseInt)
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* [。更换()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace)
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* [。重复()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat)
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## ![:clipboard:](https://forum.freecodecamp.com/images/emoji/emoji_one/clipboard.png?v=3 ":剪贴板:")捐款说明:
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* ![:warning:](https://forum.freecodecamp.com/images/emoji/emoji_one/warning.png?v=3 ":警告:") **请勿**添加与任何现有解决方案类似的解决方案。如果您认为它**_相似但更好_** ,那么尝试合并(或替换)现有的类似解决方案。
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* 添加解决方案的说明。
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* 将解决方案分为以下类别之一 - **基本** , **中级**和**高级** 。 ![:traffic_light:](https://forum.freecodecamp.com/images/emoji/emoji_one/traffic_light.png?v=3 ":红绿灯:")
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* 如果您添加了任何**相关的主要内容,**请仅添加您的用户名。 ( ![:warning:](https://forum.freecodecamp.com/images/emoji/emoji_one/warning.png?v=3 ":警告:") **_不要_** _删除任何现有的用户名_ )
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> 看到![:point_right:](https://forum.freecodecamp.com/images/emoji/emoji_one/point_right.png?v=3 ":point_right:") [**`Wiki Challenge Solution Template`**](http://forum.freecodecamp.com/t/algorithm-article-template/14272)供参考。
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