我添加了两个以字符串形式输入的二进制数，并使用此method将它们的和以二进制形式输出。
有关让我的代码工作的任何想法吗？
def add(a,b):
a = list(a)
b = list(b)
equalize(a,b)
sum_check(a,b,ctr)
out = " ".join(str(x) for x in sum_check(a,b,ctr))
out.replace(" ","")
print(out)
def equalize(a,b):
if len(a) > len(b):
for i in range(0, len(a)-len(b)):
b.insert(0,'0')
elif len(a) < len(b):
for i in range(0, len(b)-len(a)):
a.insert(0,'0')
def sum_check(a,b):
out = []
ctr = 0
def sum(a, b):
if ctr > 0:
if a[-1] + b[-1] == 2:
out.append('1')
elif a[-1] + b[-1] == 0:
out.append('1')
ctr -= 1
else: # a[-1] + b[-1] = 1
out.append('0')
ctr -= 1
else: # ctr = 0
if a[-1] + b[-1] == 2:
out.append('1')
ctr += 1
elif a[-1] + b[-1] == 0:
out.append('0')
else: # a[-1] + b[-1] = 1
out.append('1')
for i in range(len(a)):
if i == 0:
sum(a,b)
else:
new_a = a[:-1]
new_b = b[:-1]
sum(new_a, new_b)
return out
答案 0 :(得分：0)
你的算法并不是很简单（我会完全重写它;我也至少会给函数和变量更明确的名称，以便以后更快地理解算法，就像ctr无论如何应该明确地与携带有关等）。
然而，在这里，它是，纠正和工作。我在更改内容的地方插入了一些注释（算法中的错误或python编程错误）。
我已将def sum(...)放在sum_check之外，这样更清晰。
虽然您应该知道sum是python builtin，但您应该找到另一个名称，以避免丢失命名空间中的内置内容（但我猜您的算法仅用于培训目的，因为您可以通过直接对二进制数进行操作来替换所有这些：
$ python3
Python 3.4.3 (default, Oct 14 2015, 20:28:29)
[GCC 4.8.4] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> a = int('10110', 2)
>>> b = int('1011100', 2)
>>> bin(a + b)
'0b1110010'
>>>
）。
另外，我留下了一些调试print语句，以帮助查看会发生什么。再加上一个检查所有案例的例子。
最后注意：它首先崩溃，因为您在定义它之前使用了ctr。这是第一件要纠正的事情。
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
def equalize(a, b):
if len(a) > len(b):
for i in range(0, len(a) - len(b)):
b.insert(0, 0)
elif len(a) < len(b):
for i in range(0, len(b) - len(a)):
a.insert(0, 0)
return a, b
def sum(a, b, ctr):
out = ''
print('\n'.join(['-------', 'Working on:', str(a), str(b)]))
if ctr > 0:
if a[-1] + b[-1] == 2:
out = '1'
print('Case 1')
elif a[-1] + b[-1] == 0:
out = '1'
ctr -= 1
print('Case 2')
else: # a[-1] + b[-1] = 1
out = '0'
print('Case 3')
# ctr -= 1 (wrong)
else: # ctr = 0
if a[-1] + b[-1] == 2:
out = '0' # '1' was wrong
ctr += 1
print('Case 4')
elif a[-1] + b[-1] == 0:
out = '0'
print('Case 5')
else: # a[-1] + b[-1] = 1
out = '1'
print('Case 6')
print('Sum will return: ' + str(out)
+ ' and carry: ' + str(ctr) + '\n-------')
return out, ctr
def sum_check(a, b):
ctr = 0
out = []
n = len(a)
for i in range(n):
if i != 0:
# You were always giving the same a and b to feed sum().
# In order to change them, as it's not advised to iterate over a
# changing list (maybe not even possible), I stored the desired
# length in a number n. Then, I assign new values to a and b.
a = a[:-1]
b = b[:-1]
new_out, ctr = sum(a, b, ctr)
out.append(new_out)
print('Current out: ' + str(out) + ' and carry: ' + str(ctr))
return out
def add(a, b):
a = [int(x) for x in a]
b = [int(x) for x in b]
# You need to return the new contents of a and b, otherwise you'll keep
# them as they were before the call to equalize()
a, b = equalize(a, b)
print('\n'.join(['Equalized: ', str(a), str(b)]))
# On next line, [::-1] reverses the result (your algorithm returns a
# result to be read from right to left)
print('Result: ' + ''.join(sum_check(a, b)[::-1]))
add('10110', '1011100')
输出：
Equalized:
[0, 0, 1, 0, 1, 1, 0]
[1, 0, 1, 1, 1, 0, 0]
-------
Working on:
[0, 0, 1, 0, 1, 1, 0]
[1, 0, 1, 1, 1, 0, 0]
Case 5
Sum will return: 0 and carry: 0
-------
Current out: ['0'] and carry: 0
-------
Working on:
[0, 0, 1, 0, 1, 1]
[1, 0, 1, 1, 1, 0]
Case 6
Sum will return: 1 and carry: 0
-------
Current out: ['0', '1'] and carry: 0
-------
Working on:
[0, 0, 1, 0, 1]
[1, 0, 1, 1, 1]
Case 4
Sum will return: 0 and carry: 1
-------
Current out: ['0', '1', '0'] and carry: 1
-------
Working on:
[0, 0, 1, 0]
[1, 0, 1, 1]
Case 3
Sum will return: 0 and carry: 1
-------
Current out: ['0', '1', '0', '0'] and carry: 1
-------
Working on:
[0, 0, 1]
[1, 0, 1]
Case 1
Sum will return: 1 and carry: 1
-------
Current out: ['0', '1', '0', '0', '1'] and carry: 1
-------
Working on:
[0, 0]
[1, 0]
Case 2
Sum will return: 1 and carry: 0
-------
Current out: ['0', '1', '0', '0', '1', '1'] and carry: 0
-------
Working on:
[0]
[1]
Case 6
Sum will return: 1 and carry: 0
-------
Current out: ['0', '1', '0', '0', '1', '1', '1'] and carry: 0
Result: 1110010