Prime Number of Set Bits in Binary Representation
Given two integers left and right, return the count of numbers in the inclusive range [left, right] 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 1's present when written in binary.
For example, 21 written in binary is 10101, which has 3 set bits.
Example 1:
Input: left = 6, right = 10
Output: 4
Explanation:
6 -> 110 (2 set bits, 2 is prime)
7 -> 111 (3 set bits, 3 is prime)
8 -> 1000 (1 set bit, 1 is not prime)
9 -> 1001 (2 set bits, 2 is prime)
10 -> 1010 (2 set bits, 2 is prime)
4 numbers have a prime number of set bits.
Example 2:
Input: left = 10, right = 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)
5 numbers have a prime number of set bits.
Constraints:
1 <= left <= right <= 106
0 <= right - left <= 104
思路一：暴力求解
public int countPrimeSetBits(int left, int right) {
int count = 0;
for (int i = left; i <= right; i++) {
if (isNumBitCountsPrime(i)) {
count++;
}
}
return count;
}
public boolean isNumBitCountsPrime(int num) {
int count = 0;
while (num > 0) {
if ((num & 1) == 1) {
count++;
}
num >>= 1;
}
if (count == 1) {
return false;
}
for (int i = 2; i < count; i++) {
if (count % i == 0) {
return false;
}
}
return true;
}