204. Count Primes

Question

Given an integer n, return the number of prime numbers that are strictly less than n.

Example 1:

Input: n = 10
Output: 4
Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.

Example 2:

Input: n = 0
Output: 0

Example 3:

Input: n = 1
Output: 0

Constraints:

• 0 <= n <= 5 * 106

Algorithm

The basic thought is that you can use brute force to check each number less than n is a prime;

Or

You can utilize a smarter way: https://leetcode.com/problems/count-primes/solution/

Check out this solution explanation.

j start from i*I is because it doesn't have to start from 2 since all those numbers is less than j and may be calculated.

j += I each time the number is skip by I because we are checking the product of i

Code

class Solution {
    public int countPrimes(int n) {
        if (n == 1 || n == 0) return 0;
        boolean[] memo = new boolean[n];
        for (int i = 2; i <= (int) Math.sqrt(n); i++) {
            if (memo[i] == false) {
                for (int j = i*i; j < n; j += i) {
                    memo[j] = true;
                }
            }
        }
        
        int res = 0;
        for (boolean m : memo) {
            if (!m) {
                res++;
            }
        }
        return res - 2;
    }
}