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

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本文介绍了一个LeetCode的解题报告，讨论了如何找出在指定范围内，二进制表示中有素数个1的整数数量。通过暴力求解方法，对每个数计算其二进制表示中的1的个数，并检查该数量是否为素数。优化策略是预先存储32以内所有素数，并使用哈希表快速判断。

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.

思路：

暴力做法：我们从L到R，依次计算每个数中有多少个bits，如果发现它有素数个bits，就更新结果。由于int是32位的，所以我们只需要检测小于32的素数即可，为了加速处理，我们将其加入到一个哈希表中，每次只需要查找bits是否已经存在于哈希表中即可。

代码：

class Solution {
public:
    int countPrimeSetBits(int L, int R) {
        unordered_set<int> primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
        int count = 0, bits = 0;
        for (int i = L; i <= R; ++i) {
            bits = 0;
            for (int n = i; n > 0; n >>= 1) {     // calcualte the count of bits in i
                bits += n & 1;
            }
            count += primes.count(bits);
        }
        return count;
    }
};