leetcode-problemset/leetcode-cn/originData/find-kth-bit-in-nth-binary-string.json

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            "title": "Find Kth Bit in Nth Binary String",
            "titleSlug": "find-kth-bit-in-nth-binary-string",
            "content": "<p>Given two positive integers <code>n</code> and <code>k</code>, the binary string <code>S<sub>n</sub></code> is formed as follows:</p>\n\n<ul>\n\t<li><code>S<sub>1</sub> = &quot;0&quot;</code></li>\n\t<li><code>S<sub>i</sub> = S<sub>i - 1</sub> + &quot;1&quot; + reverse(invert(S<sub>i - 1</sub>))</code> for <code>i &gt; 1</code></li>\n</ul>\n\n<p>Where <code>+</code> denotes the concatenation operation, <code>reverse(x)</code> returns the reversed string <code>x</code>, and <code>invert(x)</code> inverts all the bits in <code>x</code> (<code>0</code> changes to <code>1</code> and <code>1</code> changes to <code>0</code>).</p>\n\n<p>For example, the first four strings in the above sequence are:</p>\n\n<ul>\n\t<li><code>S<sub>1 </sub>= &quot;0&quot;</code></li>\n\t<li><code>S<sub>2 </sub>= &quot;0<strong>1</strong>1&quot;</code></li>\n\t<li><code>S<sub>3 </sub>= &quot;011<strong>1</strong>001&quot;</code></li>\n\t<li><code>S<sub>4</sub> = &quot;0111001<strong>1</strong>0110001&quot;</code></li>\n</ul>\n\n<p>Return <em>the</em> <code>k<sup>th</sup></code> <em>bit</em> <em>in</em> <code>S<sub>n</sub></code>. It is guaranteed that <code>k</code> is valid for the given <code>n</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 3, k = 1\n<strong>Output:</strong> &quot;0&quot;\n<strong>Explanation:</strong> S<sub>3</sub> is &quot;<strong><u>0</u></strong>111001&quot;.\nThe 1<sup>st</sup> bit is &quot;0&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 4, k = 11\n<strong>Output:</strong> &quot;1&quot;\n<strong>Explanation:</strong> S<sub>4</sub> is &quot;0111001101<strong><u>1</u></strong>0001&quot;.\nThe 11<sup>th</sup> bit is &quot;1&quot;.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 20</code></li>\n\t<li><code>1 &lt;= k &lt;= 2<sup>n</sup> - 1</code></li>\n</ul>\n",
            "translatedTitle": "找出第 N 个二进制字符串中的第 K 位",
            "translatedContent": "<p>给你两个正整数 <code>n</code> 和 <code>k</code>，二进制字符串  <code>S<sub>n</sub></code> 的形成规则如下：</p>\n\n<ul>\n\t<li><code>S<sub>1</sub> = \"0\"</code></li>\n\t<li>当 <code>i > 1</code> 时，<code>S<sub>i</sub> = S<sub>i-1</sub> + \"1\" + reverse(invert(S<sub>i-1</sub>))</code></li>\n</ul>\n\n<p>其中 <code>+</code> 表示串联操作，<code>reverse(x)</code> 返回反转 <code>x</code> 后得到的字符串，而 <code>invert(x)</code> 则会翻转 x 中的每一位（0 变为 1，而 1 变为 0）。</p>\n\n<p>例如，符合上述描述的序列的前 4 个字符串依次是：</p>\n\n<ul>\n\t<li><code>S<sub>1 </sub>= \"0\"</code></li>\n\t<li><code>S<sub>2 </sub>= \"0<strong>1</strong>1\"</code></li>\n\t<li><code>S<sub>3 </sub>= \"011<strong>1</strong>001\"</code></li>\n\t<li><code>S<sub>4</sub> = \"0111001<strong>1</strong>0110001\"</code></li>\n</ul>\n\n<p>请你返回  <code>S<sub>n</sub></code> 的 <strong>第 <code>k</code> 位字符</strong> ，题目数据保证 <code>k</code> 一定在 <code>S<sub>n</sub></code> 长度范围以内。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 3, k = 1\n<strong>输出：</strong>\"0\"\n<strong>解释：</strong>S<sub>3</sub> 为 \"<strong>0</strong>111001\"，其第 1 位为 \"0\" 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 4, k = 11\n<strong>输出：</strong>\"1\"\n<strong>解释：</strong>S<sub>4</sub> 为 \"0111001101<strong>1</strong>0001\"，其第 11 位为 \"1\" 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 1, k = 1\n<strong>输出：</strong>\"0\"\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 2, k = 3\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 <= n <= 20</code></li>\n\t<li><code>1 <= k <= 2<sup>n</sup> - 1</code></li>\n</ul>\n",
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