描述
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
题意是找到小于n的质数的个数,如果是用常规方法判定,肯定会超时,其实可以采用厄拉托斯特尼筛法,先用2筛选,把2的倍数筛掉,再用下一个质数3,把3的倍数筛掉,接下来是5,把5的倍数筛掉,不断重复,直到$\sqrt{n}$
解决方案1(C++)
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| class Solution { public: int countPrimes(int n) { bool *result = new bool[n]; result[2] = false; int count = 0; for(int i = 3; i < n; i++) { if(i % 2 == 0) { result[i] = true; }else { result[i] = false; } } for(int i = 3; i*i < n; i += 2) { if(!result[i]) { for(int j = 2; i*j < n; j++) { result[i*j] = true; } } } for(int i = 2; i < n; i++) { if(!result[i]) { count++; } } return count; } };
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