题目
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.
题目地址
讲解
这一题同样是针对二进制表示进行出题，题目的意思是计算一个数的二进制表示中有多少个 1，如果是素数个 1，结果就加一。性能的提升点应该是在 isPrime 函数的实现上。我采用的是最普通的方法，从 2 到 $\sqrt x$ 进行累加，判断 x 是否是素数，效率比较低。
java 代码
class Solution {
public int countPrimeSetBits(int L, int R) {
int result = 0;
for(int i=L;i<=R;i++){
int temp = i;
int count = 0;
while(temp>0){
if(temp%2==1){
count++;
}
temp >>= 1;
}
if(isPrime(count)){
result++;
}
}
return result;
}
private boolean isPrime(int a){
if(a<=1){
return false;
}
for(int i=2;i<=Math.sqrt(a);i++){
if(a%i==0){
return false;
}
}
return true;
}
}