LintCode 1046 · Prime Number of Set Bits in Binary Representation (求比特数和质数题)-优快云博客

文章介绍了一个问题，给定两个整数L和R，找出范围内具有质数数量位的数字个数。提供了解决方案，利用质数位数列表和位运算计算每个数的位数，统计满足条件的数字总数。

1046 · Prime Number of Set Bits in Binary Representation
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Description
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.)

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

Example
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)

解法1：
注意 R in the range [1, 10^6]，所以最大数也只有20个比特。
每个数的1数目和，应该就是数组primes中的一个。primes = {2, 3, 5, 7, 11, 13, 17, 19};

class Solution {
public:
    /**
     * @param l: an integer
     * @param r: an integer
     * @return: the count of numbers in the range [L, R] having a prime number of set bits in their binary representation
     */
    int countPrimeSetBits(int l, int r) {
        int primeBits = 0;
        vector<int> primes = {2, 3, 5, 7, 11, 13, 17, 19};
        set<int> primeSet(primes.begin(), primes.end());
        int res = 0;
        for (int i = l; i <= r; i++) {
            int numBits = calcBits(i);
            if (primeSet.find(numBits) != primeSet.end()) {
                res++;
            }
        }
        return res;
    }
private:
    int calcBits(int x) {
       int numBits = 0;
       while (x) {
           x = x & (x - 1);
           numBits++;
       }
       return numBits;
    }
};