Given two integers L and R, find the count of numbers in the range [L, R] (inclusive) having a prime number of set bits in their binary representation.

(Recall that the number of set bits an integer has is the number of 1s present when written in binary. For example, 21 written in binary is 10101 which has 3 set bits. Also, 1 is not a prime.)

Example 1:

Input: L = 6, R = 10
Output: 4
Explanation:
6 -> 110 (2 set bits, 2 is prime)
7 -> 111 (3 set bits, 3 is prime)
9 -> 1001 (2 set bits , 2 is prime)
10->1010 (2 set bits , 2 is prime)

Example 2:

Input: L = 10, R = 15
Output: 5
Explanation:
10 -> 1010 (2 set bits, 2 is prime)
11 -> 1011 (3 set bits, 3 is prime)
12 -> 1100 (2 set bits, 2 is prime)
13 -> 1101 (3 set bits, 3 is prime)
14 -> 1110 (3 set bits, 3 is prime)
15 -> 1111 (4 set bits, 4 is not prime)

Note:

1. L, R will be integers L <= R in the range [1, 10^6].
2. R - L will be at most 10000.

这道题给了我们一个整数范围[L, R]，让我们统计其中有多个整数，其二进制表示中非零位个数为质数。参考题目中的例子不难理解题意，那么博主最先想到的就是暴力搜索啊，毕竟是到Easy题嘛，可能不需要太多的技巧。我们遍历整数范围[L, R]中的每一个数字，然后先统计出所有非零位个数cnt，通过和1相与，再右移一位的方式。然后就是来判断这个cnt是否是质数，判断的方法就是就是从其平方开始，一个一个的除，如果一直到2都没有约数，那么就是质数啦，结果res累加1，参见代码如下：

解法一：

class Solution {
public:
    int countPrimeSetBits(int L, int R) {
        int res = 0;
        for (int i = L; i <= R; ++i) {
            int t = i, cnt = 0;
            while (t > 0) {
                if (t & 1 == 1) ++cnt;
                t >>= 1;
            }
            bool succ = true;
            for (int j = sqrt(cnt); j > 1; --j) {
                if (cnt % j == 0) {
                    succ = false; break;
                }
            }
            if (succ && cnt != 1) ++res;
        }
        return res;
    }
};

好，下面我们来优化一下上面的解法，由于题目中给了数的大小范围 R <= 106 < 220，那么我们统计出来的非零位个数cnt只需要检测是否是20以内的质数即可，所以我们将20以内的质数都放入一个HashSet中，然后统计出来cnt后，直接在HashSet中查找有没有即可，参见代码如下：

解法二：

class Solution {
public:
    int countPrimeSetBits(int L, int R) {
        int res = 0;
        unordered_set<int> primes{2, 3, 5, 7, 11, 13, 17, 19};
        for (int i = L; i <= R; ++i) {
            int cnt = 0;
            for (int j = i; j > 0; j >>= 1) {
                cnt += j & 1;
            }
            res += primes.count(cnt);
        }
        return res;
    }
};

下面这种写法就更简洁啦，直接使用了C++的内置函数__builtin_popcount来快速的求出非零位的个数cnt，然后又利用到了20以内的数，只要不能被2和3的一定是质数，又可以快速判断了质数了，参见代码如下：

解法三：

class Solution {
public:
    int countPrimeSetBits(int L, int R) {
        int res = 0;
        for (int i = L; i <= R; ++i) {
            int cnt = __builtin_popcount(i);
            res += cnt < 4 ? cnt > 1 : (cnt % 2 && cnt % 3);
        }
        return res;
    }
};

类似题目：

Number of 1 Bits

参考资料：

https://leetcode.com/problems/prime-number-of-set-bits-in-binary-representation/discuss/113225/Short-C++-12-ms

https://leetcode.com/problems/prime-number-of-set-bits-in-binary-representation/discuss/113227/JavaC++-Clean-Code

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