Add 204_Count_Primes Python and Java solution based on CPP Solution

Author: qiyuangong

README.md

@@ -81,6 +81,7 @@ Also, there are open source implementations for basic data structs and algorithm
 | 186 | [Reverse Words in a String II](https://leetcode.com/problems/reverse-words-in-a-string-ii/) ♥| [Python](https://github.com/qiyuangong/leetcode/blob/master/python/186_Reverse_Words_in_a_String_II.py) | Reverse all and reverse each words |
 | 198 | [House Robber](https://leetcode.com/problems/house-robber/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/198_House_Robber.py) | f(k) = max(f(k – 2) + num[k], f(k – 1)), O(n) and O(1) |
 | 200 | [Number of Islands](https://leetcode.com/problems/number-of-islands/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/200_Number_of_Islands.py) | 1. Quick union find, O(nlogn) and O(n^2)<br>2. BFS with marks, O(n^2) and O(1) |
+| 204 | [Count Primes](https://leetcode.com/problems/count-primes/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/204_Count_Primes.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/204_Count_Primes.java) [CPP](https://github.com/qiyuangong/leetcode/blob/master/cpp/204_Count_Primes.cpp) | [Sieve of Eratosthenes](https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes#Algorithm_complexity), O(nloglogn) and O(n) |
 | 206 | [Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/206_Reverse_Linked_List.py) | 1. Stack, O(n) and O(n)<br>2. Traverse on prev and curr, then curr.next = prev, O(n) and O(1)<br>3. Recursion, O(n) and O(1) | 
 | 213 | [House Robber II](https://leetcode.com/problems/house-robber-ii/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/213_House_Robber_II.py) | f(k) = max(f(k – 2) + num[k], max(dp[0~ls-2],dp[1~ls-1], O(n) and O(1)|
 | 215 | [Kth Largest Element in an Array](https://leetcode.com/problems/kth-largest-element-in-an-array/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/215_Kth_Largest_Element_in_an_Array.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/215_Kth_Largest_Element_in_an_Array.java) | 1. Sort, O(n) and O(n)<br>2. Heap, O(nlgk) and O(n)<br>3. Quick selection, O(klgn) and O(n)|

CountingPrimes.cpp renamed to cpp/204_Count_Primes.cpp

@@ -1,6 +1,5 @@
 // Source : https://leetcode.com/problems/count-primes/
 
-
 /********************************************************************************** 
  * 
  * Description:
@@ -113,36 +112,43 @@
 #include <vector>
 using namespace std;
 
-int countPrimes(int n) {
+int countPrimes(int n)
+{
     vector<bool> isPrimer(n, true);
 
-    for(int i=2; i*i<n; i++){
-        if (isPrimer[i]){
-            for(int j=i*i; j<n; j+=i){
+    for (int i = 2; i * i < n; i++)
+    {
+        if (isPrimer[i])
+        {
+            for (int j = i * i; j < n; j += i)
+            {
                 isPrimer[j] = false;
             }
         }
     }
 
     int cnt = 0;
-    for(int i=2; i<n; i++){
-        if (isPrimer[i]) { 
+    for (int i = 2; i < n; i++)
+    {
+        if (isPrimer[i])
+        {
             //cout << i << ", ";
             cnt++;
         }
     }
     return cnt;
 }
 
-
-int main(int argc, char**argv) 
+int main(int argc, char **argv)
 {
     int n = 100;
-    if (argc>1){
+    if (argc > 1)
+    {
         n = atoi(argv[1]);
     }
-  
-    cout << endl << n << " : " << countPrimes(n) << endl;
+
+    cout << endl
+         << n << " : " << countPrimes(n) << endl;
 
     return 0;
 }

java/204_Count_Primes.java

@@ -0,0 +1,20 @@
+class Solution {
+    // ttps://en.wikipedia.org/wiki/Sieve_of_Eratosthenes#Algorithm_complexity
+    public int countPrimes(int n) {
+        boolean[] isPrime = new boolean[n];
+        int count = 0;
+        Arrays.fill(isPrime, true);
+        for (int i = 2; i < n; i++) {
+            if (i * i >= n)
+                break;
+            if (!isPrime[i])
+                continue;
+            for (int j = i * i; j < n; j += i)
+                isPrime[j] = false;
+        }
+        for (int i = 2; i < n; i++)
+            if (isPrime[i])
+                count++;
+        return count;
+    }
+}

python/204_Count Primes.py renamed to python/204_Count_Primes.py

@@ -17,4 +17,4 @@ def countPrimes(self, n):
         for i in xrange(2, n):
             if isPrime[i]:
                 count += 1
-        return count
+        return count