leetcode_Count Primes

描述：

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

思路：

判断一个数是否为质数有两种方法，一种是判断它能否被２～(int)sqrt(n)+1之间的数整除，能被整除为合数，否则为质数

但是，当n非常大的时候这种方法是非常费时间的。

另外一种改进的方法是仅用n除以2~(int)sqrt(n)+1之间的所有质数，而2~(int)sqrt(n)+1之间的质数从何而来？先计算出来

测试用例需要的小于num的所有质数的num大概为1200就能通过本题的测试用例

代码：

public int countPrimes(int n) {        n--;        if(n<2)            return 0;        if(n==2)            return 1;        int primArr[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279};        int count=1;        int temp=0;        for(int i=3;i<=n;i++)        {            count++;            temp=(int)Math.sqrt(i)+1;            for(int j=0;j<primArr.length;j++)            {                if(temp>primArr[j])                {                    if(i%primArr[j]==0)                    {                        count--;                        break;                    }                 }else                    break;            }        }        return count;    }