leetcode-problemset/leetcode-cn/originData/lexicographical-numbers.json

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            "questionId": "386",
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            "categoryTitle": "Algorithms",
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            "title": "Lexicographical Numbers",
            "titleSlug": "lexicographical-numbers",
            "content": "<p>Given an integer <code>n</code>, return all the numbers in the range <code>[1, n]</code> sorted in lexicographical order.</p>\n\n<p>You must write an algorithm that runs in&nbsp;<code>O(n)</code>&nbsp;time and uses <code>O(1)</code> extra space.&nbsp;</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<pre><strong>Input:</strong> n = 13\n<strong>Output:</strong> [1,10,11,12,13,2,3,4,5,6,7,8,9]\n</pre><p><strong class=\"example\">Example 2:</strong></p>\n<pre><strong>Input:</strong> n = 2\n<strong>Output:</strong> [1,2]\n</pre>\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "字典序排数",
            "translatedContent": "<p>给你一个整数 <code>n</code> ，按字典序返回范围 <code>[1, n]</code> 内所有整数。</p>\n\n<p>你必须设计一个时间复杂度为 <code>O(n)</code> 且使用 <code>O(1)</code> 额外空间的算法。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 13\n<strong>输出：</strong>[1,10,11,12,13,2,3,4,5,6,7,8,9]\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 2\n<strong>输出：</strong>[1,2]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>\n</ul>\n",
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                    "name": "Depth-First Search",
                    "slug": "depth-first-search",
                    "translatedName": "深度优先搜索",
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                    "lang": "C++",
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                    "code": "class Solution {\npublic:\n    vector<int> lexicalOrder(int n) {\n\n    }\n};",
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                    "code": "class Solution(object):\n    def lexicalOrder(self, n):\n        \"\"\"\n        :type n: int\n        :rtype: List[int]\n        \"\"\"",
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                    "code": "/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* lexicalOrder(int n, int* returnSize) {\n    \n}",
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