Leetcode -- Count Primes

最新推荐文章于 2022-09-27 11:10:45 发布

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本文介绍了一种高效的算法来计算小于非负整数n的所有素数的数量。通过筛除已知非素数的倍数来逐步减少候选素数集，实现了优化的素数筛选过程。该算法特别适用于大范围内的素数计数。

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题目：
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.

分析【引自：Leetcode Hint】：
这里写图片描述
We start off with a table of n numbers. Let’s look at the first number, 2. We know all multiples of 2 must not be primes, so we mark them off as non-primes. Then we look at the next number, 3. Similarly, all multiples of 3 such as 3 × 2 = 6, 3 × 3 = 9, … must not be primes, so we mark them off as well. Now we look at the next number, 4, which was already marked off. What does this tell you? Should you mark off all multiples of 4 as well?

4 is not a prime because it is divisible by 2, which means all multiples of 4 must also be divisible by 2 and were already marked off. So we can skip 4 immediately and go to the next number, 5. Now, all multiples of 5 such as 5 × 2 = 10, 5 × 3 = 15, 5 × 4 = 20, 5 × 5 = 25, … can be marked off. There is a slight optimization here, we do not need to start from 5 × 2 = 10. Where should we start marking off?

In fact, we can mark off multiples of 5 starting at 5 × 5 = 25, because 5 × 2 = 10 was already marked off by multiple of 2, similarly 5 × 3 = 15 was already marked off by multiple of 3. Therefore, if the current number is p, we can always mark off multiples of p starting at p2, then in increments of p: p2 + p, p2 + 2p, … Now what should be the terminating loop condition?

代码：

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