Python |计算范围内未设置的位数
原文:https://www . geesforgeks . org/python-count-unset-bits-range/
给定一个非负数 n 和两个值 l 和 r，问题是在 n 的二进制表示中计算 l 到 r 范围内的未设置位的数量，即从最右边的第 1 位到最右边的第 1 位计算未设置位。
示例:
Input : n = 42, l = 2, r = 5
Output : 2
(42)10 = (101010)2
There are '2' unset bits in the range 2 to 5.
Input : n = 80, l = 1, r = 4
Output : 4
对于这个问题，我们有现有的解决方案，请参考计算范围内未设置的位链接。我们可以用 Python 快速解决这个问题。方法很简单，
- Use the bin (num) function to convert decimal to binary.
- Now delete the first two characters of the output binary string, because by default, the bin function appends' 0b' to the output string as a prefix.
- Slice the string from the index (l-1) to index r and invert it, and then calculate the unset bits in the middle.
# Function to count unset bits in a range
def unsetBits(n,l,r):
# convert n into it's binary
binary = bin(n)
# remove first two characters
binary = binary[2:]
# reverse string
binary = binary[-1::-1]
# count all unset bit '0' starting from index l-1
# to r, where r is exclusive
print (len([binary[i] for i in range(l-1,r) if binary[i]=='0']))
# Driver program
if __name__ == "__main__":
n=42
l=2
r=5
unsetBits(n,l,r)
输出:
2
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