[LeetCode] 204. Count Primes 计数质数

Description:

Count the number of prime numbers less than a non-negative number, n

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References:

How Many Primes Are There?

Sieve of Eratosthenes

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

计数出小于非负整数n的质数数量。质数（prime number）又称素数，有无限个。质数定义为在大于1的自然数中，除了1和它本身以外不再有其他因数。

解法：埃拉托斯特尼筛法 Sieve of Eratosthenes

如果一个数是另一个数的倍数，那这个数肯定不是质数。利用这个性质，可以建立一个质数数组，从2开始将素数的倍数都标注为不是质数。第一轮将4、6、8等表为非质数，然后遍历到3，发现3没有被标记为非质数，则将6、9、12等标记为非质数，一直到N为止，再数一遍质数数组中有多少质数。

Java:

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public class Solution {     public int countPrimes(int n) {         boolean[] prime = new boolean[n];         Arrays.fill(prime, true);         for(int i = 2; i < n; i++){             if(prime[i]){                 // 将i的2倍、3倍、4倍...都标记为非素数                 for(int j = i * 2; j < n; j =  j + i){                     prime[j] = false;                 }             }         }         int count = 0;         for(int i = 2; i < n; i++){             if(prime[i]) count++;         }         return count;     } }

Python:

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class Solution:     # @param {integer} n     # @return {integer}     def countPrimes(self, n):         isPrime = [True] * max(n, 2)         isPrime[0], isPrime[1] = False, False         x = 2         while x * x < n:             if isPrime[x]:                 p = x * x                 while p < n:                     isPrime[p] = False                     p += x             x += 1         return sum(isPrime)

Python:

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class Solution(object):     def countPrimes(self, n):         """         :type n: int         :rtype: int         """         if n <= 2: return 0         vis = [False] * n         for i in range(2, int(n ** 0.5) + 1):             if vis[i]: continue             j = i             while j * i < n:                 vis[j * i] = True                 j += 1         ans = 0         for i in range(2, n):             if not vis[i]: ans += 1         return ans

C++:

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class Solution { public:     int countPrimes(int n) {         if(!n||n==1)  return 0;         vector<bool> isPrime(n,true);         // Loop's ending condition is i * i < n instead of i < sqrt(n)         // to avoid repeatedly calling an expensive function sqrt().         for(int i=2;i*i<n;++i)         {             if(!isPrime[i]) continue;             //填表起点i*i，如3*3，因为3*2已填，步长+i             for(int j=i*i;j<n;j+=i)             {                 isPrime[j]=false;             }         }         int count=0;         for(int i=2;i<n;++i)         {             if(isPrime[i])  ++count;         }         return count;     } };

　　

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