leetcode-problemset/leetcode-cn/originData/integer-break.json

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            "questionId": "343",
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            "categoryTitle": "Algorithms",
            "boundTopicId": 1110,
            "title": "Integer Break",
            "titleSlug": "integer-break",
            "content": "<p>Given an integer <code>n</code>, break it into the sum of <code>k</code> <strong>positive integers</strong>, where <code>k &gt;= 2</code>, and maximize the product of those integers.</p>\n\n<p>Return <em>the maximum product you can get</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 2\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> 2 = 1 + 1, 1 &times; 1 = 1.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 10\n<strong>Output:</strong> 36\n<strong>Explanation:</strong> 10 = 3 + 3 + 4, 3 &times; 3 &times; 4 = 36.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 58</code></li>\n</ul>\n",
            "translatedTitle": "整数拆分",
            "translatedContent": "<p>给定一个正整数&nbsp;<code>n</code>&nbsp;，将其拆分为 <code>k</code> 个 <strong>正整数</strong> 的和（&nbsp;<code>k &gt;= 2</code>&nbsp;），并使这些整数的乘积最大化。</p>\n\n<p>返回 <em>你可以获得的最大乘积</em>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入: </strong>n = 2\n<strong>输出: </strong>1\n<strong>解释: </strong>2 = 1 + 1, 1 × 1 = 1。</pre>\n\n<p><strong>示例&nbsp;2:</strong></p>\n\n<pre>\n<strong>输入: </strong>n = 10\n<strong>输出: </strong>36\n<strong>解释: </strong>10 = 3 + 3 + 4, 3 ×&nbsp;3 ×&nbsp;4 = 36。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 58</code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    int integerBreak(int n) {\n\n    }\n};",
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                    "code": "class Solution(object):\n    def integerBreak(self, n):\n        \"\"\"\n        :type n: int\n        :rtype: int\n        \"\"\"",
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                "There is a simple O(n) solution to this problem.",
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