762. Prime Number of Set Bits in Binary Representation

762. Prime Number of Set Bits in Binary Representation

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题目

Leetcode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.

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解决

1.利用位运算来数1的个数
【n为[L, R]的数字个数，len为R的二进制的位数】

• 时间复杂度：O(n * m)
• 空间复杂度：O(1)

// First Solution(Runtime: 39ms)
class Solution {
public:
    int countPrimeSetBits(int L, int R) {
        set<int> primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
        int result = 0;
        for (int i = L; i <= R; i++) {
            // determine every number in [L, R]
            int bits = 0;
            for (int b = i; b; b >>= 1) { // shift operation
                // determine if the end is 1
                bits += b & 1;
            }
            result += primes.count(bits);
        }
        return result;
    }
};

2.利用bitset来数1的个数
【n为[L, R]的数字个数，len为R的二进制的位数】

• 时间复杂度：O(n * m)
• 空间复杂度：O(1)

虽然复杂度没有太大的变化，但是利用bitset可以简化十进制转换为二进制。
并且bitset中有函数count()可以统计其中1的个数。

// Second Solution(Runtime: 21ms)
class Solution {
public:
    int countPrimeSetBits(int L, int R) {
        set<int> primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
        int result = 0;
        for (int i = L; i <= R; i++) {
            bitset<100> b(i);
            result += primes.count(b.count());
        }
        return result;
    }
};