leetcode 204题求素数个数

Description:

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

提示晒数法：

http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

https://primes.utm.edu/howmany.html

别人的代码：

int countPrimes(int n) {
if (n<=2) return 0;
vector<bool> passed(n, false);
int sum = 1;
int upper = sqrt(n);
for (int i=3; i<n; i+=2) {
if (!passed[i]) {
sum++;
//avoid overflow
if (i>upper) continue;
for (int j=i*i; j<n; j+=i) {
passed[j] = true;
}
}
}
return sum;
}

我的代码：

// countprime.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"

#include <iostream>
#include <math.h>
#include<vector>
using namespace std;

int countPrimes1(int n)
{
int temp = 0;
if (2 >= n) return 0;

bool* primes = new bool
;
for (int i = 2; i < n; ++i)
primes[i] = true;

int sqr = (int)(sqrt((double)(n - 1)));
for (int i = 2; i <= sqr; ++i)
{
if (primes[i])
{
temp++;
for (int j = i * i; j < n; j += i)
primes[j] = false;
}
}

int sum = 0;
for (int i = 2; i < n; ++i)
sum += (primes[i]) ? 1 : 0;

/*cout<<temp;*/
delete[] primes;

return sum;
}

int countPrimes(int n)
{
if (n<=2)return 0;

int sum = 0;
int sqr = (int)(sqrt((double)(n - 1)));

vector<bool> prime(n,0);

for(int i = 2; i < n; ++i)
prime[i] = 1;

for(int i =2; i <= sqr; ++i)
{
if(prime[i]==1)
{
//sum++;
for(int j = i*i; j < n; j = j+i)
prime[j] = 0;
}

}

for(int i = 2;i < n; ++i)
sum += prime[i] ? 1 : 0;

return sum;
}

int _tmain(int argc, _TCHAR* argv[])
{

cout<<countPrimes(5)<<endl;

cout<<countPrimes1(3)<<endl;

getchar();
return 0;
}

超时代码：

int countPrimes(int n)
{
int sum = 0;
for(int i = 0 ;i<=n;i=i+2)
{
int j = 2;
int temp = sqrt(i);
for(;j<=temp;j=j+1)
{
if((i%temp)==0)
{break;}
sum++;
}
}

return sum;
}

纪念一下首次AC