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[n]; 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