[Leetcode] 762. Prime Number of Set Bits in Binary Representation
2207 ワード
Description:
Given two integers
(Recall that the number of set bits an integer has is the number of
Example 1:
Example 2:
Note:
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; } }
それぞれ10進法を用いてバイナリを回転し,質量数判断の2つのアルゴリズムを用いた.
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 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; } }
それぞれ10進法を用いてバイナリを回転し,質量数判断の2つのアルゴリズムを用いた.