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, Rwill be integersL <= Rin the range[1, 10^6].R - Lwill be at most 10000.
给定两个整数L和R，找到在其二进制表示中具有设置位素数的范围[L，R]（包含）范围内的数字的计数。 （回想一下，一个整数所设置的位数是用二进制写的时候存在的1的个数，例如，用二进制写的21是10101，它有3个设置位，而且1不是素数。
/*** @param {number} L* @param {number} R* @return {number}*/var countPrimeSetBits = function (L, R) {let primes = new Set([2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]);let res = 0;while (L <= R) {let str = L.toString(2);let match = str.match(/1/g);if (match && primes.has(match.length)) {res++;}L++;}return res;};let res = countPrimeSetBits(10, 15);console.log(res);
