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 integersL <= R
in the range[1, 10^6]
.R - L
will 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);