(easy) Leetcode 204.Count Primes

Description:

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

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

Resolution: A natural number greater than 1, the natural number can be divided by 1 and its own, then the natural number is called Prime.

Method One: Brute force, Time complexity O (n^2)

The code is as follows:

public class Solution {public    int countprimes (int n) {        if (n<=2) return 0;        int num=0;        for (int i=2;i<n;i++)            if (IsPrime (i))                num++;        return num;     } public boolean isprime (int i) {for    (int j=2;j<i;j++) {          //or for (int j=2;j<=i/2;j++), or for (int j=2;j*j <i;j++);        if (i%j==0)            return false;    }    return true;   }}

　　

Run Result: timeout, time complexity O (n^2)

Method 2: Multiples of prime numbers are excluded.

The code is as follows:

public class Solution {public     int countprimes (int n) {       if (n<=2) return 0;       Boolean []isprime=new boolean[n];       int sum=0;       for (int i=2;i<n;i++) {           isprime[i]=true;       }       for (int i=2;i<n;i++) {           if (Isprime[i]) {for               (int j=2;j*i<n;j++) {                   isprime[j*i]=false;}}       } for       (int i=2;i<n;i++) {           if (isprime[i]==true)             sum++;       }       return sum;    } }

　　

Run Result: Time complexity O (n).

(easy) Leetcode 204.Count Primes