LeetCode-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.)

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:

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

题意：给定一个范围[L, R]，遍历这个区间的每一个元素，计算这个元素的二进制表示中1的个数是否为一个素数；返回这个区间满足此条件的元素的个数；

解法：只要遍历这个区间，对每个元素转换为二进制表示后计算其1的个数是否为素数即可；因为1的个数可能相同，我们用一个集合来表示为当前已计算过的为素数的元素，这样如果下次有相同的元素，就不需要再重复判断是否为素数了；否则，需要重新判断是否为素数；

Java

class Solution {
    public int countPrimeSetBits(int L, int R) {
        Set<Integer> primes = new HashSet<>();
        int result = 0;
        
        for (int i = L; i <= R; i++) {
            int setBit = setBits(i);
            if (primes.contains(setBit)) result++;
            else if (isPrime(setBit)) {
                primes.add(setBit);
                result++;
            }
        }
        
        return result;
    }
    
    private int setBits(int num) {
        int cnt = 0;
        while (num > 0) {
            cnt = num % 2 == 1 ? cnt + 1 : cnt;
            num /= 2;
        }
        return cnt;
    }
    
    private boolean isPrime(int num) {
        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                return false;
            }
        }
        return num == 1 ? false : true;
    }
}