[Python/Java]finding prime num in n range template

[ZhangCheng-zh]

erichsen

MX = 10 ** 6 + 1
primeList = []
is_prime = [True] * MX
for i in range(2, MX):
    if is_prime[i]:
        primeList.append(i)
        for j in range(i * i, MX, i):
            is_prime[j] = False
primeList.extend((MX, MX))  # 保证下面下标不会越界

function erichsen (n) {
    const isPrimeList = new Array(n + 1).fill(1)
      for (i = 2; i <= n / i; i++) {
          if (isPrimeList[i] === 1) {
              // delete i Multiple number
              for (let j = i * i; j <= n; j += i) {
                  isPrimeList[j] = 0
              }
          }
      }
  
      let count = 0
      for (let i = 2; i <= n; i++) {
          if (isPrimeList[i] === 1) count++
      }
      return count
  }

euler

class Solution {
        private final static int MX = 1000000;
    private final static ArrayList<Integer> primes = new ArrayList<Integer>();

    static {
        boolean[] noPrime = new boolean[MX + 1];
        for (int i = 2; i <= MX; i++) {
            if (!noPrime[i]) {
                primes.add(i);
            }

            for (int p: primes) {
                if (i * p > MX) break;
                noPrime[i * p] = true;
                if (i % p == 0) {
                    break;
                }
            }
        }
        // optional suffix, to avoid overflow
        primes.add(MX + 1);
        primes.add(MX + 1);
    }

    // todo
}

function euler(n) {
 const isPrimeList = new Array(n + 1).fill(1)
 const primes = []
 
 let k = 0
 let count = 0
 for (let i = 2; i <= n; i++) {
  if(isPrimeList[i] === 1) { primes[k++] = i , count++ }
  for (let j = 0; primes[j] * i <= n; j++) {
     // 每个质数都和i相乘 得到合数
     isPrimeList[primes[j] * i] =  0
    // 在删除的过程中，只通过数的最小质因数筛数。
     if ( i % primes[j] === 0) break
  }
 }
}