[Leetcode] 762. Prime Number of Set Bits in Binary Representation 解题报告

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 

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.
思路：

暴力做法：我们从L到R，依次计算每个数中有多少个bits，如果发现它有素数个bits，就更新结果。由于int是32位的，所以我们只需要检测小于32的素数即可，为了加速处理，我们将其加入到一个哈希表中，每次只需要查找bits是否已经存在于哈希表中即可。

代码：class Solution {
public:
int countPrimeSetBits(int L, int R) {
unordered_set<int> primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
int count = 0, bits = 0;
for (int i = L; i <= R; ++i) {
bits = 0;
for (int n = i; n > 0; n >>= 1) { // calcualte the count of bits in i
bits += n & 1;
}
count += primes.count(bits);
}
return count;
}
};