leetcode 762. Prime Number of Set Bits in Binary Representation

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.

解法1:

直接暴力

复制代码
class Solution(object):
    def countPrimeSetBits(self, L, R):
        """
        :type L: int
        :type R: int
        :rtype: int
        """
        # for echo num:
        #    count bits in num and judge if it is prime        
        prime_nums = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}
        
        def count_1bits(n):
            ans = 0
            while n:
                ans += 1
                n = n & (n-1)
            return ans
        
        ans = 0
        for n in range(L, R+1):
            bits = count_1bits(n)
            if bits in prime_nums:
                ans += 1
        return ans
复制代码

解法2:使用dp,比较巧妙!因为 数字num中1的个数=num/2中1的个数+num末尾数字是否为1

虽然会说超时,但还是值得掌握的。

复制代码
class Solution(object):
    def countPrimeSetBits(self, L, R):
        """
        :type L: int
        :type R: int
        :rtype: int
        """      
        prime_nums = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}
        
        def count_bits(n):
            bits = [0]*(n+1)
            for i in range(1, n+1):
                bits[i] = bits[i>>1] + (i&1)
            return bits

        ans = 0
        bits = count_bits(R)
        for n in range(L, R+1):
            if bits[n] in prime_nums:
                ans += 1                
        return ans
复制代码

 

posted @   bonelee  阅读(407)  评论(0编辑  收藏  举报
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