1046 · Prime Number of Set Bits in Binary Representation
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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.)
1.L, R will be integers L <= R in the range [1, 10^6].
2.R - L will be at most 10000.
Example
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)
解法1：
注意 R in the range [1, 10^6]，所以最大数也只有20个比特。
每个数的1数目和，应该就是数组primes中的一个。primes = {2, 3, 5, 7, 11, 13, 17, 19};
class Solution {
public:
/**
* @param l: an integer
* @param r: an integer
* @return: the count of numbers in the range [L, R] having a prime number of set bits in their binary representation
*/
int countPrimeSetBits(int l, int r) {
int primeBits = 0;
vector<int> primes = {2, 3, 5, 7, 11, 13, 17, 19};
set<int> primeSet(primes.begin(), primes.end());
int res = 0;
for (int i = l; i <= r; i++) {
int numBits = calcBits(i);
if (primeSet.find(numBits) != primeSet.end()) {
res++;
}
}
return res;
}
private:
int calcBits(int x) {
int numBits = 0;
while (x) {
x = x & (x - 1);
numBits++;
}
return numBits;
}
};