Count Primes--Leetcode

Source: Internet
Author: User


Count the number of prime numbers less than a non-negative number, n.

Basic idea: Sieve method

1, 2 is the prime number, sieve out to 2 as a factor. i.e. 2 * 2, 2*3, 2*4,2*5

2, find the next number that is not sieved, such as 3. Then sift out the number with a factor of 3.

3, repeat step 2.

Time Complexity of O (n)

Class Solution {public:    int countprimes (int n) {        vector<int> sieve (n, true);        int count = 0;        for (int i=2; i<n; i++) {            if (Sieve[i]) {                ++count;                for (int j=i+i; j<n; j+=i) {                    Sieve[j] = false;        }}} return count;    }};

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Count Primes--Leetcode

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