leetcode讲解762.Prime Number of Set Bits in Binary Representation-优快云博客

本文探讨了一个特定的二进制位数问题，即在给定范围内，找出所有具有质数个1的二进制表示的整数。通过实例解释了如何计算二进制位上的1的数量，并判断其是否为质数。提供了Java代码实现，包括一个isPrime函数用于判断质数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 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, R will be integers L <= R in the range [1, 10^6].
R - L will be at most 10000.

题目地址

讲解

这一题同样是针对二进制表示进行出题，题目的意思是计算一个数的二进制表示中有多少个1，如果是素数个1，结果就加一。性能的提升点应该是在 isPrime 函数的实现上。我采用的是最普通的方法，从2到$\sqrt x$进行累加，判断x是否是素数，效率比较低。

java代码

class Solution {
    public int countPrimeSetBits(int L, int R) {
        int result = 0;
        for(int i=L;i<=R;i++){
            int temp = i;
            int count = 0;
            while(temp>0){
                if(temp%2==1){
                    count++;
                }
                temp >>= 1;
            }
            if(isPrime(count)){
                result++;
            }
        }
        return result;
    }
    
    private boolean isPrime(int a){
        if(a<=1){
            return false;
        }
        for(int i=2;i<=Math.sqrt(a);i++){
            if(a%i==0){
                return false;
            }
        }
        return true;
    }
}