LeetCode – Count Primes (Java)

 

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

Java Solution 1

This solution exceeds time limit.

public int countPrimes(int n) {
    n = n-1;
 
    ArrayList<Integer> primes = new ArrayList<Integer>();
 
    if(n<=1) 
        return 0;
    if(n==2)
        return 1;
    if(n==3)
        return 2;
 
    primes.add(2);
    primes.add(3);
 
    for(int i=4; i<=n; i++){
        boolean isPrime = true;
        for(int p: primes){
            int m = i%p;
            if(m==0){
                isPrime = false;
                break;
            }
        }
 
        if(isPrime){
            primes.add(i);
        }
    }
 
    return primes.size();
}

Java Solution 2

This solution is the implementation of Sieve of Eratosthenes.

public int countPrimes(int n) {
	if (n <= 2)
		return 0;
 
	// init an array to track prime numbers
	boolean[] primes = new boolean[n];
	for (int i = 2; i < n; i++)
		primes[i] = true;
 
	for (int i = 2; i <= Math.sqrt(n - 1); i++) {
	// or for (int i = 2; i <= n-1; i++) {
		if (primes[i]) {
			for (int j = i + i; j < n; j += i)
				primes[j] = false;
		}
	}
 
	int count = 0;
	for (int i = 2; i < n; i++) {
		if (primes[i])
			count++;
	}
 
	return count;
}

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  1. Vj on 2015-4-29

    We can avoid first and last for loop in above code. Refer below code,

    public static void countPrimes()

    {

    int n = 20;

    bool[] nums = new bool[n+1];

    int nonPrimeCounter = 0;

    for (int i = 2; i <= 7; i++)

    {

    if (nums[i] == false)

    {

    for (int j = i+i; j <= n; j = j + i)

    {

    if (j%i == 0)

    {

    if (nums[j] == false)

    {

    nums[j] = true;

    nonPrimeCounter++;

    }

    }

    }

    }

    }

    int primCounter = n – nonPrimeCounter;

    }

  2. RMFOREVER123 on 2016-1-21

    public int countPrimes(int n) {
    if (n <= 2)
    return 0;

    // init an array to track prime numbers
    boolean[] primes = new boolean[n];
    for (int i = 2; i < n; i++)
    primes[i] = true;

    for (int i = 2; i <= Math.sqrt(n - 1); i++) {
    // or for (int i = 2; i <= n-1; i++) {
    if (primes[i]) {

    // j=i*i not i+i
    for (int j = i * i; j < n; j += i)
    primes[j] = false;
    }
    }

    int count = 0;
    for (int i = 2; i < n; i++) {
    if (primes[i])
    count++;
    }

    return count;
    }

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