update

leetcode-cn/originData/the-kth-factor-of-n.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你两个正整数&nbsp;<code>n</code> 和&nbsp;<code>k</code>&nbsp;。</p>\n\n<p>如果正整数 <code>i</code> 满足 <code>n % i == 0</code> ，那么我们就说正整数 <code>i</code> 是整数 <code>n</code>&nbsp;的因子。</p>\n\n<p>考虑整数 <code>n</code>&nbsp;的所有因子，将它们 <strong>升序排列</strong>&nbsp;。请你返回第 <code>k</code>&nbsp;个因子。如果 <code>n</code>&nbsp;的因子数少于 <code>k</code>&nbsp;，请你返回 <strong>-1</strong>&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 = 12, k = 3\n<strong>输出：</strong>3\n<strong>解释：</strong>因子列表包括 [1, 2, 3, 4, 6, 12]，第 3 个因子是 3 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 7, k = 2\n<strong>输出：</strong>7\n<strong>解释：</strong>因子列表包括 [1, 7] ，第 2 个因子是 7 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 4, k = 4\n<strong>输出：</strong>-1\n<strong>解释：</strong>因子列表包括 [1, 2, 4] ，只有 3 个因子，所以我们应该返回 -1 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= n &lt;= 1000</code></li>\n</ul>\n",
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             "hints": [
                 "The factors of n will be always in the range [1, n].",
                 "Keep a list of all factors sorted.  Loop i from 1 to n and add i if n % i == 0. Return the kth factor if it exist in this list."