762. Prime Number of Set Bits in Binary Representation #

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:

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

题目大意 #

给定两个整数 L 和 R ，找到闭区间 [L, R] 范围内，计算置位位数为质数的整数个数。（注意，计算置位代表二进制表示中1的个数。例如 21 的二进制表示 10101 有 3 个计算置位。还有，1 不是质数。）

注意:

• L, R 是 L <= R 且在 [1, 10^6] 中的整数。
• R - L 的最大值为 10000。

解题思路 #

• 题目给出 [L, R] 区间，在这个区间内的每个整数的二进制表示中 1 的个数如果是素数，那么最终结果就加一，问最终结果是多少？这一题是一个组合题，判断一个数的二进制位有多少位 1，是第 191 题。题目中限定了区间最大不超过 10^6 ，所以 1 的位数最大是 19 位，也就是说素数最大就是 19 。那么素数可以有限枚举出来。最后按照题目的意思累积结果就可以了。

代码 #

package leetcode

import "math/bits"

func countPrimeSetBits(L int, R int) int {
	counter := 0
	for i := L; i <= R; i++ {
		if isPrime(bits.OnesCount(uint(i))) {
			counter++
		}
	}
	return counter
}

func isPrime(x int) bool {
	return x == 2 || x == 3 || x == 5 || x == 7 || x == 11 || x == 13 || x == 17 || x == 19
}