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 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 integersL <= R
in the range[1, 10^6]
.R - L
will be at most 10000.
解法1:
直接暴力
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
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。
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
标签:
leetcode
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