leetcode-problemset/leetcode-cn/originData/integer-to-roman.json

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            "title": "Integer to Roman",
            "titleSlug": "integer-to-roman",
            "content": "<p>Seven different symbols represent Roman numerals with the following values:</p>\n\n<table>\n\t<thead>\n\t\t<tr>\n\t\t\t<th>Symbol</th>\n\t\t\t<th>Value</th>\n\t\t</tr>\n\t</thead>\n\t<tbody>\n\t\t<tr>\n\t\t\t<td>I</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>V</td>\n\t\t\t<td>5</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>X</td>\n\t\t\t<td>10</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>L</td>\n\t\t\t<td>50</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>C</td>\n\t\t\t<td>100</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>D</td>\n\t\t\t<td>500</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>M</td>\n\t\t\t<td>1000</td>\n\t\t</tr>\n\t</tbody>\n</table>\n\n<p>Roman numerals are formed by appending&nbsp;the conversions of&nbsp;decimal place values&nbsp;from highest to lowest. Converting a decimal place value into a Roman numeral has the following rules:</p>\n\n<ul>\n\t<li>If the value does not start with 4 or&nbsp;9, select the symbol of the maximal value that can be subtracted from the input, append that symbol to the result, subtract its value, and convert the remainder to a Roman numeral.</li>\n\t<li>If the value starts with 4 or 9 use the&nbsp;<strong>subtractive form</strong>&nbsp;representing&nbsp;one symbol subtracted from the following symbol, for example,&nbsp;4 is 1 (<code>I</code>) less than 5 (<code>V</code>): <code>IV</code>&nbsp;and 9 is 1 (<code>I</code>) less than 10 (<code>X</code>): <code>IX</code>.&nbsp;Only the following subtractive forms are used: 4 (<code>IV</code>), 9 (<code>IX</code>),&nbsp;40 (<code>XL</code>), 90 (<code>XC</code>), 400 (<code>CD</code>) and 900 (<code>CM</code>).</li>\n\t<li>Only powers of 10 (<code>I</code>, <code>X</code>, <code>C</code>, <code>M</code>) can be appended consecutively at most 3 times to represent multiples of 10. You cannot append 5&nbsp;(<code>V</code>), 50 (<code>L</code>), or 500 (<code>D</code>) multiple times. If you need to append a symbol&nbsp;4 times&nbsp;use the <strong>subtractive form</strong>.</li>\n</ul>\n\n<p>Given an integer, convert it to a Roman numeral.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">num = 3749</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">&quot;MMMDCCXLIX&quot;</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<pre>\n3000 = MMM as 1000 (M) + 1000 (M) + 1000 (M)\n 700 = DCC as 500 (D) + 100 (C) + 100 (C)\n  40 = XL as 10 (X) less of 50 (L)\n   9 = IX as 1 (I) less of 10 (X)\nNote: 49 is not 1 (I) less of 50 (L) because the conversion is based on decimal places\n</pre>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">num = 58</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">&quot;LVIII&quot;</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<pre>\n50 = L\n 8 = VIII\n</pre>\n</div>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">num = 1994</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">&quot;MCMXCIV&quot;</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<pre>\n1000 = M\n 900 = CM\n  90 = XC\n   4 = IV\n</pre>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= num &lt;= 3999</code></li>\n</ul>\n",
            "translatedTitle": "整数转罗马数字",
            "translatedContent": "<p>七个不同的符号代表罗马数字，其值如下：</p>\n\n<table>\n\t<thead>\n\t\t<tr>\n\t\t\t<th>符号</th>\n\t\t\t<th>值</th>\n\t\t</tr>\n\t</thead>\n\t<tbody>\n\t\t<tr>\n\t\t\t<td>I</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>V</td>\n\t\t\t<td>5</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>X</td>\n\t\t\t<td>10</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>L</td>\n\t\t\t<td>50</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>C</td>\n\t\t\t<td>100</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>D</td>\n\t\t\t<td>500</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>M</td>\n\t\t\t<td>1000</td>\n\t\t</tr>\n\t</tbody>\n</table>\n\n<p>罗马数字是通过添加从最高到最低的小数位值的转换而形成的。将小数位值转换为罗马数字有以下规则：</p>\n\n<ul>\n\t<li>如果该值不是以 4 或 9 开头，请选择可以从输入中减去的最大值的符号，将该符号附加到结果，减去其值，然后将其余部分转换为罗马数字。</li>\n\t<li>如果该值以 4 或 9 开头，使用 <strong>减法形式</strong>，表示从以下符号中减去一个符号，例如&nbsp;4 是 5 (<code>V</code>) 减 1 (<code>I</code>): <code>IV</code>&nbsp;，9 是 10 (<code>X</code>) 减&nbsp;1 (<code>I</code>)：<code>IX</code>。仅使用以下减法形式：4 (<code>IV</code>)，9 (<code>IX</code>)，40 (<code>XL</code>)，90 (<code>XC</code>)，400 (<code>CD</code>) 和&nbsp;900 (<code>CM</code>)。</li>\n\t<li>只有 10 的次方（<code>I</code>, <code>X</code>, <code>C</code>, <code>M</code>）最多可以连续附加 3 次以代表 10 的倍数。你不能多次附加&nbsp;5&nbsp;(<code>V</code>)，50 (<code>L</code>) 或 500 (<code>D</code>)。如果需要将符号附加4次，请使用 <strong>减法形式</strong>。</li>\n</ul>\n\n<p>给定一个整数，将其转换为罗马数字。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong><span class=\"example-io\">num = 3749</span></p>\n\n<p><strong>输出：</strong>&nbsp;<span class=\"example-io\">\"MMMDCCXLIX\"</span></p>\n\n<p><strong>解释：</strong></p>\n\n<pre>\n3000 = MMM 由于 1000 (M) + 1000 (M) + 1000 (M)\n 700 = DCC 由于 500 (D) + 100 (C) + 100 (C)\n  40 = XL 由于 50 (L) 减 10 (X)\n   9 = IX 由于 10 (X) 减 1 (I)\n注意：49 不是 50 (L) 减 1 (I) 因为转换是基于小数位\n</pre>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong><span class=\"example-io\">num = 58</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">\"LVIII\"</span></p>\n\n<p><strong>解释：</strong></p>\n\n<pre>\n50 = L\n 8 = VIII\n</pre>\n</div>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong><span class=\"example-io\">num = 1994</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">\"MCMXCIV\"</span></p>\n\n<p><strong>解释：</strong></p>\n\n<pre>\n1000 = M\n 900 = CM\n  90 = XC\n   4 = IV\n</pre>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= num &lt;= 3999</code></li>\n</ul>\n",
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\\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.11<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap \\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"http:\\/\\/www.grantjenks.com\\/docs\\/sortedcontainers\\/\\\" target=\\\"_blank\\\">sortedcontainers<\\/a> \\u5e93\\u3002<\\/p>\"],\"scala\":[\"Scala\",\"<p>\\u7248\\u672c\\uff1a<code>Scala 3.3.1<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"<p>\\u7248\\u672c\\uff1a<code>Kotlin 1.9.0<\\/code><\\/p>\\r\\n\\r\\n<p>\\u6211\\u4eec\\u4f7f\\u7528\\u7684\\u662f JetBrains \\u63d0\\u4f9b\\u7684 experimental compiler\\u3002\\u5982\\u679c\\u60a8\\u8ba4\\u4e3a\\u60a8\\u9047\\u5230\\u4e86\\u7f16\\u8bd1\\u5668\\u76f8\\u5173\\u7684\\u95ee\\u9898\\uff0c\\u8bf7\\u5411\\u6211\\u4eec\\u53cd\\u9988<\\/p>\"],\"rust\":[\"Rust\",\"<p>\\u7248\\u672c\\uff1a<code>rust 1.79.0<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 crates.io \\u7684 <a href=\\\"https:\\/\\/crates.io\\/crates\\/rand\\\" target=\\\"_blank\\\">rand<\\/a> \\u548c <a href=\\\"https:\\/\\/crates.io\\/crates\\/regex\\\" target=\\\"_blank\\\">regex<\\/a><\\/p>\"],\"php\":[\"PHP\",\"<p><code>PHP 8.2<\\/code>.<\\/p>\\r\\n\\r\\n<p>With bcmath module.<\\/p>\"],\"typescript\":[\"TypeScript\",\"<p>TypeScript 5.1.6<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2022<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/blob\\/v5\\/README.md\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.4.0<\\/a>\\uff0c<a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/v4.2.3\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.3<\\/a> \\u4ee5\\u53ca <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/deque\\/tree\\/v1.0.4\\\" target=\\\"_blank\\\"> datastructures-js\\/deque@1.0.4<\\/a>\\u3002<\\/p>\"],\"racket\":[\"Racket\",\"<p><a href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.15<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 26\"],\"elixir\":[\"Elixir\",\"Elixir 1.17 with Erlang\\/OTP 26\"],\"dart\":[\"Dart\",\"<p>Dart 3.2\\u3002\\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/pub.dev\\/packages\\/collection\\/versions\\/1.18.0\\\" target=\\\"_blank\\\">collection<\\/a> \\u5305<\\/p>\\r\\n\\r\\n<p>\\u60a8\\u7684\\u4ee3\\u7801\\u5c06\\u4f1a\\u88ab\\u4e0d\\u7f16\\u8bd1\\u76f4\\u63a5\\u8fd0\\u884c<\\/p>\"],\"cangjie\":[\"Cangjie\",\"<p>\\u7248\\u672c\\uff1a0.53.11 (cjnative)<\\/p>\\r\\n\\r\\n<p>\\u7f16\\u8bd1\\u53c2\\u6570\\uff1a<code>-O2 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