Leetcode 204 Count Primes (prime number Count) (*)

translation

计算小于一个非负整数n的质数的个数。

Original

thenumberofanumber, n.

Analysis

This problem encountered before, at that time is the most stupid method, and now there is no good idea, and just the topic has a hint, I opened. The tip of the topic is a one to give out, I also on one after the whole point opened, feel good failure ...

 Public int CountPrimes(intN) {intCount =0; for(inti =1; I < n; i++) {if(IsPrime (i)) count++; }returnCount;}Private Boolean IsPrime(intNUM) {if(Num <=1)return false;//Loop ' s ending condition is i * I <= num instead of i <= sqrt (num)   //To avoid repeatedly calling an expensive function sqrt ().    for(inti =2; I * i <= num; i++) {if(num% i = =0)return false; }return true;}

ofisofthetotothattheis surprisingly simple.

 Public intCountPrimes (intN) {Boolean[] IsPrime =New Boolean[n]; for(inti =2; I < n; i++) {Isprime[i] =true; }//Loop ' s ending condition is I * I < n instead of I < sqrt (n)   //To avoid repeatedly calling an expensive function sqrt ().    for(inti =2; I * i < n; i++) {if(!isprime[i])Continue; for(intj = i * I; J < N; J + = i) {Isprime[j] =false; }   }int Count=0; for(inti =2; I < n; i++) {if(Isprime[i])Count++; }return Count;}

The above are Leetcode original ...

Translate the above code into the following:

Class Solution { Public:BOOL IsPrime(intNUM) {if(Num <=1)return false; for(inti =2; I*i <= num; ++i) {if(num%i = =0)return false; }return true; }intCountPrimes (intN) {BOOL*isprime =New BOOL[n]; for(inti =2; I < n; ++i) {Isprime[i] =true; } for(inti =2; I*i < n; ++i) {if(!isprime[i])Continue; for(intj = i*i; J < N; J + = i) {Isprime[j] =false; }        }intCount =0; for(inti =2; I < n; ++i) {if(Isprime[i]) count++; }returnCount }};

Excerpt some code:

Class Solution { Public:int CountPrimes(intN) {Switch(n) { Case 0: Case 1: Case 2:return 0; Case 3:return 1; Case 4: Case 5:return 2; Case 6: Case 7:return 3; Case 8: Case 9: Case Ten: Case  One:return 4; Case  A: Case  -:return 5; Case  -: Case  the:return 6; Case 10000:return 1229; Case 499979:return 41537; Case 999983:return 78497; Case 1500000:return 114155; }    }};

intCountPrimes (intN) {if(--n <2)return 0;intM = (n +1)/2,Count= m, k, u = (sqrt (n)-1)/2; BOOL Notprime[m] = {0}; for(inti =1; I <= u;i++)if(!notprime[i]) for(k = (i+1)*2*i; K < M;k + = i*2+1)if(!notprime[k]) {Notprime[k] =true;Count--; }return Count;}

Leetcode 204 Count Primes (prime number Count) (*)