[LeetCode] Prime Number of Set Bits in Binary Representation

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.

找出从L到R中数字的二进制中1的个数是否是质数的数字。并统计L到R中这样的数字的个数。

思路：

1、判断一个数是否为质数

2、判断一个 数字的二进制位中有多少个1

3、循环判断L~R中的每一个数，并统计

class Solution {
public:
    int countPrimeSetBits(int L, int R) {
        int res = 0;
        for (int i = L; i <= R; i++) {
            int cnt = countBits(i);
            if (isPrime(cnt))
                res++;
        }
        return res;
    }

    int countBits(int num) {
        int res = 0;
        while (num) {
            if (num & 1)
                res++;
            num = num >> 1;
        }
        return res;
    }

    bool isPrime(int num) {
        if (num <= 3)
            return num > 1;
        for (int i = 2; i <= sqrt(num); i++) {
            if (num % i == 0)
                return false;
        }
        return true;
    }
};
// 23 ms

原文地址：https://www.cnblogs.com/immjc/p/8288808.html