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LeetCode-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.

Analysis:

See hint in leetcode.

Solution:

public class Solution {    public int countPrimes(int n) {        if (n<=2) return 0;                boolean[] marked = new boolean[n];                for (int i=3;i*i<n;i+=2)            if (!marked[i]){                // i*i+(2*k+1)*i = i*(i+1+2k) is an even value because (i+1) is even.                // We can skip all such number by increasing value by 2*i every time.                for (int value = http://www.mamicode.com/i*i;valuei){                    marked[value]=true;                }            }                    int count = 1;        for (int i=3;i<n;i+=2)            if (!marked[i]){                count++;            }        return count;    }}

 

LeetCode-Count Primes