import java.util.Arrays;
/**
* <p>给定整数 <code>n</code> ，返回 <em>所有小于非负整数 <code>n</code> 的质数的数量</em> 。</p>
*
* <p> </p>
*
* <p><strong>示例 1：</strong></p>
*
* <pre>
* <strong>输入：</strong>n = 10
* <strong>输出：</strong>4
* <strong>解释：</strong>小于 10 的质数一共有 4 个, 它们是 2, 3, 5, 7 。
* </pre>
*
* <p><strong>示例 2：</strong></p>
*
* <pre>
* <strong>输入：</strong>n = 0
* <strong>输出：</strong>0
* </pre>
*
* <p><strong>示例 3：</strong></p>
*
* <pre>
* <strong>输入：</strong>n = 1
* <strong>输出</strong>：0
* </pre>
*
* <p> </p>
*
* <p><strong>提示：</strong></p>
*
* <ul>
* <li><code>0 <= n <= 5 * 10<sup>6</sup></code></li>
* </ul>
* <div><div>Related Topics</div><div><li>数组</li><li>数学</li><li>枚举</li><li>数论</li></div></div><br><div><li>👍 929</li><li>👎 0</li></div>
*/
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
//避免全部扫描
// 12 = 2 × 6
//12 = 3 × 4
//12 = sqrt(12) × sqrt(12)
//12 = 4 × 3
//12 = 6 × 2
public int countPrimes(int n) {
boolean[] isPrimes = new boolean[n + 1];
Arrays.fill(isPrimes, true);
for (int i = 2; i < n; i++) {
if (isPrimes[i]) {
for (int j = 2 * i; j <= n; j += i) {
isPrimes[j] = false;
}
}
}
int count = 0;
for (int i = 2; i < n; i++) {
if (isPrimes[i]) {
count++;
}
}
return count;
}
}
//leetcode submit region end(Prohibit modification and deletion)不会，我可以学；落后，我可以追赶；跌倒，我可以站起来！
