JZ15 二进制中1的个数
二进制中1的个数
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编程打卡第三天：
解法1：分情况讨论，用bin()
def NumberOf1(self , n: int) -> int:
# write code here
result = 0
result_neg = 0
if n >= 0:
binary_n = bin(n)
for strs in str(binary_n):
if strs == '1':
result += 1
return result
if n < 0:
binary_n_neg = bin(abs(n))[2:]
for strs in str(binary_n_neg):
if strs == "0":
result_neg += 1
result = 32- result_neg
return result
解法2： 反码：n & 0xffffffff
def NumberOf1(self , n: int) -> int:
# write code here
# -*- coding:utf-8 -*-
# write code here
if n < 0:
n = n & 0xffffffff
binNum = bin(n)[2:]
arr = str(binNum)
res = 0
for ch in arr:
if int(ch) == 1:
res += 1
return res
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