Prime Number of Set Bits in Binary Representation

问题：

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 integers

L <= R

in the range

[1, 10^6]

.

R - L

will be at most 10000.

解决：

【题意】给一个数字区间范围，判断这个区间的数字的二进制形式中，1的个数为质数的数字共计多少个。

① 计算数的二进制表示中1的个数，然后判断这个个数是否为素数。

class Solution { //24ms
public int countPrimeSetBits(int L, int R) {
int res = 0;
for (int i = L;i <= R;i ++){
if (isPrime(Integer.bitCount(i))){
res ++;
}
}
return res;
}
public boolean isPrime(int n){
if (n == 1) return false;
if (n == 2 || n == 3) return true;
for (int i = 2;i <= Math.sqrt(n);i ++){
if (n % i == 0){
return false;
}
}
return true;
}
}

class Solution { //57ms
public int countPrimeSetBits(int L, int R) {
int res = 0;
for (int i = L;i <= R;i ++){
int count = countBits(i);
if (isPrime(count)){
res ++;
}
}
return res;
}
public int countBits(int n){
int count = 0;
do {
if ((n & 1) == 1){
count ++;
}
n = n >> 1;
}while (n != 0);
return count;
}
public boolean isPrime(int n){
boolean res = true;
if (n == 1) return false;
if (n == 2 || n == 3) return true;
for (int i = 2;i <= Math.sqrt(n);i ++){
if (n % i == 0){
res = false;
break;
}
}
return res;
}
}