leetcode-problemset/leetcode-cn/originData/first-bad-version.json

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            "title": "First Bad Version",
            "titleSlug": "first-bad-version",
            "content": "<p>You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.</p>\n\n<p>Suppose you have <code>n</code> versions <code>[1, 2, ..., n]</code> and you want to find out the first bad one, which causes all the following ones to be bad.</p>\n\n<p>You are given an API <code>bool isBadVersion(version)</code> which returns whether <code>version</code> is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 5, bad = 4\n<strong>Output:</strong> 4\n<strong>Explanation:</strong>\ncall isBadVersion(3) -&gt; false\ncall isBadVersion(5)&nbsp;-&gt; true\ncall isBadVersion(4)&nbsp;-&gt; true\nThen 4 is the first bad version.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 1, bad = 1\n<strong>Output:</strong> 1\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= bad &lt;= n &lt;= 2<sup>31</sup> - 1</code></li>\n</ul>\n",
            "translatedTitle": "第一个错误的版本",
            "translatedContent": "<p>你是产品经理，目前正在带领一个团队开发新的产品。不幸的是，你的产品的最新版本没有通过质量检测。由于每个版本都是基于之前的版本开发的，所以错误的版本之后的所有版本都是错的。</p>\n\n<p>假设你有 <code>n</code> 个版本 <code>[1, 2, ..., n]</code>，你想找出导致之后所有版本出错的第一个错误的版本。</p>\n\n<p>你可以通过调用 <code>bool isBadVersion(version)</code> 接口来判断版本号 <code>version</code> 是否在单元测试中出错。实现一个函数来查找第一个错误的版本。你应该尽量减少对调用 API 的次数。</p>\n \n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 5, bad = 4\n<strong>输出：</strong>4\n<strong>解释：</strong>\n<code>调用 isBadVersion(3) -> false \n调用 isBadVersion(5) -> true \n调用 isBadVersion(4) -> true</code>\n<code>所以，4 是第一个错误的版本。</code>\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 1, bad = 1\n<strong>输出：</strong>1\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= bad <= n <= 2<sup>31</sup> - 1</code></li>\n</ul>\n",
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                    "code": "// The API isBadVersion is defined for you.\n// bool isBadVersion(int version);\n\nclass Solution {\npublic:\n    int firstBadVersion(int n) {\n        \n    }\n};",
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