feat: add solutions to leetcode problem: No.0204

See https://leetcode-cn.com/problems/count-primes

Author: yanglbme

README.md

@@ -100,6 +100,7 @@
 1. [数组中数字出现的次数 II](/lcof/面试题56%20-%20II.%20数组中数字出现的次数%20II/README.md)
 1. [错误的集合](/solution/0600-0699/0645.Set%20Mismatch/README.md)
 1. [二进制中 1 的个数](/lcof/面试题15.%20二进制中1的个数/README.md)
+1. [计数质数](/solution/0200-0299/0204.Count%20Primes/README.md)
 
 ### 栈和队列
 

README_EN.md

@@ -95,6 +95,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 ### Math
 
 1. [Set Mismatch](/solution/0600-0699/0645.Set%20Mismatch/README_EN.md)
+1. [Count Primes](/solution/0200-0299/0204.Count%20Primes/README_EN.md)
 
 ### Stack & Queue
 

solution/0200-0299/0204.Count Primes/README.md

@@ -18,22 +18,57 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+如果 x 是质数，那么大于 x 的 x 的倍数 2x,3x,… 一定不是质数，因此我们可以从这里入手。
+
+我们设 `primes[i]` 表示数 i 是不是质数，如果是质数则为 true，否则为 false。从小到大遍历每个数，如果这个数为质数，则将其所有的倍数都标记为合数（除了该质数本身），即 false，这样在运行结束的时候我们即能知道质数的个数。
+
+对于一个质数 x，我们从 2x 开始标记其实是冗余的，应该直接从 x⋅x 开始标记，因为 2x,3x,… 这些数一定在 x 之前就被其他数的倍数标记过了，例如 2 的所有倍数，3 的所有倍数等。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def countPrimes(self, n: int) -> int:
+        if n < 2:
+            return 0
+        res = 0
+        primes = [True for _ in range(n)]
+        for i in range(2, n):
+            if primes[i]:
+                res += 1
+                for j in range(i * i, n, i):
+                    primes[j] = False
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int countPrimes(int n) {
+        if (n < 2) return 0;
+        boolean[] primes = new boolean[n];
+        Arrays.fill(primes, true);
+        int res = 0;
+        for (int i = 2; i < n; ++i) {
+            if (primes[i]) {
+                ++res;
+                if ((long) i * i < n) {
+                    for (int j = i * i; j < n; j += i) {
+                        primes[j] = false;
+                    }
+                }
+            }
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0200-0299/0204.Count Primes/README_EN.md

@@ -25,13 +25,42 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def countPrimes(self, n: int) -> int:
+        if n < 2:
+            return 0
+        res = 0
+        primes = [True for _ in range(n)]
+        for i in range(2, n):
+            if primes[i]:
+                res += 1
+                for j in range(i * i, n, i):
+                    primes[j] = False
+        return res
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int countPrimes(int n) {
+        if (n < 2) return 0;
+        boolean[] primes = new boolean[n];
+        Arrays.fill(primes, true);
+        int res = 0;
+        for (int i = 2; i < n; ++i) {
+            if (primes[i]) {
+                ++res;
+                if ((long) i * i < n) {
+                    for (int j = i * i; j < n; j += i) {
+                        primes[j] = false;
+                    }
+                }
+            }
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0200-0299/0204.Count Primes/Solution.java

@@ -1,25 +1,19 @@
-public class Solution {
+class Solution {
     public int countPrimes(int n) {
-
-		boolean[] b = new boolean[n];
-
-		for (int i = 2; i * i < n; i++) {
-			if (b[i] == false) {
-				for (int j = i; i * j < n; j++) {
-					b[i * j] = true;
-				}
-			}
-		}
-
-		int c = 0;
-		for (int i = 2; i < n; i++) {
-			if (b[i] == false) {
-				c++;
-			}
-		}
-
-		return c;
-	
-        
+        if (n < 2) return 0;
+        boolean[] primes = new boolean[n];
+        Arrays.fill(primes, true);
+        int res = 0;
+        for (int i = 2; i < n; ++i) {
+            if (primes[i]) {
+                ++res;
+                if ((long) i * i < n) {
+                    for (int j = i * i; j < n; j += i) {
+                        primes[j] = false;
+                    }
+                }
+            }
+        }
+        return res;
     }
 }

solution/0200-0299/0204.Count Primes/Solution.py

@@ -0,0 +1,12 @@
+class Solution:
+    def countPrimes(self, n: int) -> int:
+        if n < 2:
+            return 0
+        res = 0
+        primes = [True for _ in range(n)]
+        for i in range(2, n):
+            if primes[i]:
+                res += 1
+                for j in range(i * i, n, i):
+                    primes[j] = False
+        return res