Count the number of prime numbers less than a non-negative number, n.
Example:
Input: 10 Output: 4 Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.
Approach #1: C++.
class Solution {
public:
int countPrimes(int n) {
vector<int> matrix(n+5, 1);
helper(n, matrix);
int ans = 0;
if (n <= 1) return 0;
for (int j = 2; j < n; ++j)
if (matrix[j] == 1)
ans++;
return ans;
}
void helper(int n, vector<int>& matrix) {
for (int i = 2; i*i < n; ++i) {
if (matrix[i] == 1)
for (int j = i+i; j <= n; j += i) {
matrix[j] = 0;
}
}
}
};
Approach #2: Java.
class Solution {
public int countPrimes(int n) {
boolean[] notPrime = new boolean[n];
int count = 0;
for (int i = 2; i < n; ++i) {
if (notPrime[i] == false) {
count++;
for (int j = 2; i*j < n; ++j) {
notPrime[i*j] = true;
}
}
}
return count;
}
}
Approach #3: Python.
class Solution:
def countPrimes(self, n):
"""
:type n: int
:rtype: int
"""
if n < 3:
return 0
primes = [True] * n
primes[0] = primes[1] = False
for i in range(2, int(n**0.5)+1):
if primes[i]:
primes[i*i : n : i] = [False] * len(primes[i*i : n : i])
return sum(primes)
| Time Submitted | Status | Runtime | Language |
|---|---|---|---|
| a few seconds ago | Accepted | 240 ms | python3 |
| 3 minutes ago | Accepted | 20 ms | java |
| 7 minutes ago | Accepted | 24 ms | cpp |