leetcode204 --- Elsieve sieve and linear sieve

Essiere sieve

We consider the fact that if $x$ is a prime number, then greater than$x$ ofxxMultiples of$x$$2x$,$3x$ ,… must not be a prime number, so we can start from here.

We are $isPrime\left[i\right]$ display number$Is\; i$ a prime number, if it is a prime number, it is$1$ , otherwise$0$ . Traversing each number from small to large, if the number is a prime number, mark all its multiples as composite numbers (except the prime number itself), that is,$0$ , so that we can know the number of prime numbers at the end of the run.

You can continue to optimize here, because for a prime number $x$ , if we start from 2 x 2xas mentioned above$2\; The\; start\; mark\; ofx$ is actually redundant, it should start directly from$x\cdot x$ starts marking because$2x,3x,...$ these numbers must have been previously marked by multiples of other numbers, such as2 2All multiples of$2$$All\; multiples\; of\; 3$ etc.

class Solution {
    
    
public:
    int countPrimes(int n) {
    
    
        vector<int> isPrime(n,1);
        int res = 0;
        for(long long i=2;i<n;i++) {
    
    
            if(isPrime[i]) {
    
    
                ++res;
                for(long long j=i*i; j<n; j+=i) {
    
    
                    isPrime[j] = 0;
                }
            }
        }
        return res;
    }
};

Linear Sieve

The Esperanto sieve has the behavior of repeated marking. For example, 45 will be marked repeatedly by 3 and 5. The linear sieve can avoid this situation. The linear sieve is also called the Euler sieve. The Euler sieve needs to maintain a prime number table. For 2 ≤ $2\le i<n$ , when traversing to the integer$i$ , perform the following operations,

1, Ruo $i$ is a prime number, ie$i$ is not marked as a composite number, then$i$ join the prime number table

2. Traverse the list of prime numbers from small to large, and for each prime number $prime$ ，当 $i\ast prime<When\; n$ , perform the following operations:

​ a、将 $i\ast prime$ is marked as a composite number

​b, Ruu $i$ can be$prime$ is divisible, then end traversing the list of prime numbers

The principle here is that if if $i$ can be$prime$ divisible, then record$prim{e}^{\text{'}}$ is greater thanprime primeThe prime number of$p$$r$$im$$e$$y=x\ast prim{e}^{\prime }$ ，则有 $y=\frac{x}{prime}\ast prim{e}^{\prime }\ast prime$ , when traversing to $\frac{x}{prime} * prime' $,$y$ will be primed$prime$ mark, due to$prime<prime\text{'}$ , so when$\frac{x}{prime}$is an integer, i.e. $xmodprime==0$ , this integer can be$prim{e}^{\text{'}}$ mark, so$y$ will be labeled by its smallest prime factor.

Since the Euler sieve ensures that each composite number is labeled by its smallest prime factor, it ensures that each composite number is labeled only once.

class Solution {
    
    
public:
    int countPrimes(int n) {
    
    
        vector<int> isPrime(n, 1);
        vector<int> Prime;
        for (int i = 2; i < n; ++i) {
    
    
            if (isPrime[i]){
    
    
                Prime.push_back(i);
            } 
                
            for(auto j:Prime) {
    
    
                if ((long long)i * j < n) {
    
    
                    isPrime[i*j] = 0;
                } else {
    
    
                    break;
                }
                if (i % j == 0) break;
            }

        }
        return Prime.size();
    }
};

However, what appears to be fewer linear sieve operations actually takes longer to run because more operations are performed