LeetCode one question per day (60) 204. Counting prime numbers (tabulation method: determining prime numbers)

204. Counting prime numbers

---

---

Method 1: Enumeration

class Solution {
    
    
public:
    bool isPrime(int x) {
    
    
        for (int i = 2; i * i <= x; ++i) {
    
    
            if (x % i == 0) {
    
    
                return false;
            }
        }
        return true;
    }

    int countPrimes(int n) {
    
    
        int ans = 0;
        for (int i = 2; i < n; ++i) {
    
    
            ans += isPrime(i);
        }
        return ans;
    }
};

---

Method 2: Tabulation ( 重点)

1. First set all numbers to prime numbers (marked as 1);
2. From the first prime number 2 to n traverse, if the current number is prime, then result+1, and set the multiples of prime numbers to not prime numbers (marked as 0);

There are two sets of codes below. K is optimized starting from i. It is also possible to start from 2, but this is a bit repetitive (for example, a number x=2*i, this number is a multiple of i, which is also a multiple of 2, then This x has been set when i=2).

for(int k=i;(long long)i*k<n;k++){
    
    
     isprime[i*k]=0;
}  

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

---

The weather is a bit cold these past two days, I feel like catching a cold