8/27 每日一題(檢查二進制中1個數目是否為質數)

 Given two integers left and right, return the count of numbers in the inclusive range [left, right] 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.

 

Example 1:

Input: left = 6, right = 10
Output: 4
Explanation:
6  -> 110 (2 set bits, 2 is prime)
7  -> 111 (3 set bits, 3 is prime)
8  -> 1000 (1 set bit, 1 is not prime)
9  -> 1001 (2 set bits, 2 is prime)
10 -> 1010 (2 set bits, 2 is prime)
4 numbers have a prime number of set bits.

Example 2:

Input: left = 10, right = 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)
5 numbers have a prime number of set bits.

 

Constraints:

• 1 <= left <= right <= 106
• 0 <= right - left <= 104

class Solution:

    #檢查質數

    def is_prime(self,n):

        if n<=1:

            return False

        for i in range(2,int(n**0.5)+1):

            if n % i == 0:

                return False

        return True

   

    def countPrimeSetBits(self, left: int, right: int) -> int:

        res=0 #計算幾個質數

        for i in range(left,right+1):

            bits=bin(i)[2:].count('1')

            if self.is_prime(bits):

                res+=1

        return res

---------------------------

class Solution:
    def countPrimeSetBits(self, left: int, right: int) -> int:
        p={2,3,5,7,11,13,17,19}
        c=0
        for i in range(left,right+1):
            if i.bit_count() in p:
                c+=1
        return c     

------------------------

class Solution:
    def countPrimeSetBits(self, left: int, right: int) -> int:
        def calnumofbit(num):
            re = 0
            while num:
                if num&1:
                    re += 1
                num = num >> 1
            return re
        re = 0
        isprime = set([2,3,5,7,11,13,17,19,23])
        for n in range(left, right+1):
            num = n.bit_count()
            if num in isprime:
                re += 1
        return re