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leetcode-cn/originData/count-primes.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给定整数 <code>n</code> ，返回 <em>所有小于非负整数&nbsp;<code>n</code>&nbsp;的质数的数量</em> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 10\n<strong>输出：</strong>4\n<strong>解释：</strong>小于 10 的质数一共有 4 个, 它们是 2, 3, 5, 7 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 0\n<strong>输出：</strong>0\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 1\n<strong>输出</strong>：0\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= n &lt;= 5 * 10<sup>6</sup></code></li>\n</ul>\n",
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                 "<p>Let's start with a <i>isPrime</i> function. To determine if a number is prime, we need to check if it is not divisible by any number less than <i>n</i>. The runtime complexity of <i>isPrime</i> function would be O(<i>n</i>) and hence counting the total prime numbers up to <i>n</i> would be O(<i>n</i><sup>2</sup>). Could we do better?</p>",
                 "<p>As we know the number must not be divisible by any number > <i>n</i> / 2, we can immediately cut the total iterations half by dividing only up to <i>n</i> / 2. Could we still do better?</p>",