README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0200-0299/0204.Count%20Primes/README_EN.md	Array Math Enumeration Number Theory

204. Count Primes

中文文档

Description

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

Solutions

Solution 1

Python3

class Solution:
    def countPrimes(self, n: int) -> int:
        primes = [True] * n
        ans = 0
        for i in range(2, n):
            if primes[i]:
                ans += 1
                for j in range(i + i, n, i):
                    primes[j] = False
        return ans

Java

class Solution {
    public int countPrimes(int n) {
        boolean[] primes = new boolean[n];
        Arrays.fill(primes, true);
        int ans = 0;
        for (int i = 2; i < n; ++i) {
            if (primes[i]) {
                ++ans;
                for (int j = i + i; j < n; j += i) {
                    primes[j] = false;
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countPrimes(int n) {
        vector<bool> primes(n, true);
        int ans = 0;
        for (int i = 2; i < n; ++i) {
            if (primes[i]) {
                ++ans;
                for (int j = i; j < n; j += i) primes[j] = false;
            }
        }
        return ans;
    }
};

Go

func countPrimes(n int) int {
	primes := make([]bool, n)
	for i := range primes {
		primes[i] = true
	}
	ans := 0
	for i := 2; i < n; i++ {
		if primes[i] {
			ans++
			for j := i + i; j < n; j += i {
				primes[j] = false
			}
		}
	}
	return ans
}

JavaScript

/**
 * @param {number} n
 * @return {number}
 */
var countPrimes = function (n) {
    let primes = new Array(n).fill(true);
    let ans = 0;
    for (let i = 2; i < n; ++i) {
        if (primes[i]) {
            ++ans;
            for (let j = i + i; j < n; j += i) {
                primes[j] = false;
            }
        }
    }
    return ans;
};

C#

public class Solution {
    public int CountPrimes(int n) {
        var notPrimes = new bool[n];
        int ans = 0;
        for (int i = 2; i < n; ++i)
        {
            if (!notPrimes[i])
            {
                ++ans;
                for (int j = i + i; j < n; j += i)
                {
                    notPrimes[j] = true;
                }
            }
        }
        return ans;
    }
}