leetcode-dp-204

import java.util.Arrays;

/**
 * <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>
 *
 * <p>&nbsp;</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>&nbsp;</p>
 *
 * <p><strong>提示：</strong></p>
 *
 * <ul>
 * <li><code>0 &lt;= n &lt;= 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)