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 1
s 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 integersL <= R
in the range[1, 10^6]
.R - L
will be at most 10000.
Tags: Bit Manipulation
求在所给范围的数中,其二进制形式的1的个数为素数的数,共有多少个。
题目所给的最大范围为[1, 10^6]
,因为10^6
<2^20
,所以可以判定所有范围内的数的二进制形式中,1
的个数不会超过20,20以内的素数仅有2, 3, 5, 7, 11, 13, 17, 19。我们直接循环遍历范围内的数,计算其二进制形式中1
的个数即可。
Java:
class Solution {
public int countPrimeSetBits(int L, int R) {
int sum = 0;
for (int i = L; i <= R; i++) {
if (isPrime(Integer.bitCount(i))) {
sum++;
}
}
return sum;
}
private boolean isPrime(int num) {
return num == 2 || num == 3 || num == 5 ||
num == 7 || num == 11 || num == 13 ||
num == 17 || num == 19;
}
}
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