二进制到十进制使用递归，python

Question

需要帮助将二进制转换为十进制，使用递归。

到目前为止，我有：

（2 * INT（S [0]））+ INT（S [1]）

用碱的情况下为当s == 0和s == 1 。

我不知道如何递归传递，以便函数将通过输入中的所有1和0。

Answer 1

其基本思想是挑出字符串的最后一个字符并将其转换为数字，然后将其乘以2的适当幂。我已经为您评论了代码。

# we need to keep track of the current string, 
# the power of two, and the total (decimal) 
def placeToInt (str, pow, total): 
    # if the length of the string is one, 
    # we won't call the function anymore 
    if (len(str) == 1): 
     # return the number, 0 or 1, in the string 
     # times 2 raised to the current power, 
     # plus the already accumulated total 
     return int(str) * (2 ** pow) + total 
    else: 
     # grab the last digit, a 0 or 1 
     num = int(str[-1:]) 
     # the representation in binary is 2 raised to the given power, 
     # times the number (0 or 1) 
     # add this to the total 
     total += (num * (2 ** pow)) 
     # return, since the string has more digits 
     return placeToInt(str[:-1], pow + 1, total) 

# test case 
# appropriately returns 21 
print(placeToInt("10101", 0, 0)) 

现在，让我们手动通过它，让您明白为什么这会起作用。

# n = 101 (in binary 
# this can also be represented as 1*(2^2) + 0*(2^1) + 1*(2^0) 
# alternatively, since there are three digits in this binary number 
# 1*(2^(n-1)) + 0*(2^(n-2)) + 1*(2^(n-3)) 

那么这是什么意思？那么，最右边的数字是1或0的2次幂的零。换句话说，它要么增加1或0。第二个最右边的数字呢？它要么增加0或2。下一个？ 0或4.看到图案？

让我们写的伪代码：

let n = input, in binary 
total = 0 
power of 2 = 0 
while n has a length: 
    lastDigit = last digit of n 
    add (2^pow)*lastDigit to the current total 

因为我们开始与电力和共0，你就会明白为什么这个工程。

Answer 2

def IntegerConvert(num, base): 
    if num == 0: 
     return 0 
    else: 
     IntegerConvert.sum += pow(10, IntegerConvert.counter)*(num % base) 
     IntegerConvert.counter += 1 
     IntegerConvert(num/base, base) 
    return IntegerConvert.sum 
IntegerConvert.counter = 0 
IntegerConvert.sum = 0 
print IntegerConvert(10, 2)