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, Rwill be integersL <= Rin the range[1, 10^6]. -
R - Lwill be at most 10000.
Solution
Iterative, time O(n), space O(1)
本题目可以拆分成几道题目：
- count one bits
- isPrime
由于L和R的范围已给定，isPrime可以用枚举来优化。
class Solution {
public static final Set<Integer> primes
= new HashSet<>(Arrays.asList(2, 3, 5, 7, 11, 13, 17, 19));
public int countPrimeSetBits(int L, int R) {
int count = 0;
for (int i = L; i <= R; ++i) {
if (isPrimeSetBits(i)) {
++count;
}
}
return count;
}
public boolean isPrimeSetBits(int n) {
int oneBits = 0;
while (n > 0) {
++oneBits;
n &= n - 1;
}
return primes.contains(oneBits);
}
}
