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

最新推荐文章于 2021-12-21 00:03:43 发布

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本文介绍了一个算法问题：在给定范围内找出二进制表示中1的个数为质数的整数数量。通过将整数转换为二进制并统计1的数量，再判断该数量是否为质数来解决此问题。

Description:

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.

class Solution {
public int countPrimeSetBits(int L, int R) {
int result = 0;
for (int i = L; i <=R; i++){
   String a = Integer.toBinaryString(i);
int sum = 0;
boolean decide = true;

for (int j = 0; j < a.length(); j++) {
char c = a.charAt(j);
if (c == '1')
sum++;
}
if(sum == 2)
decide = true;
       else if(sum < 2 || sum % 2 == 0)
decide = false;
       for(int k=3; k<=Math.sqrt(sum); k+=2){
           if(sum % k == 0)
               decide = false;

       }
if (decide == true)
result = result + 1;
}
return result;
}
}