二進制到十進制使用遞歸，python

Question

需要幫助將二進制轉換爲十進制，使用遞歸。

到目前爲止，我有：

（2 * INT（S [0]））+ INT（S [1]）

用鹼的情況下爲當s == 0和s == 1 。

我不知道如何遞歸傳遞，以便函數將通過輸入中的所有1和0。

Answer 1

其基本思想是挑出字符串的最後一個字符並將其轉換爲數字，然後將其乘以2的適當冪。我已經爲您評論了代碼。

# 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)) 

現在，讓我們手動通過它，讓您明白爲什麼這會起作用。

# 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)) 

那麼這是什麼意思？那麼，最右邊的數字是1或0的2次冪的零。換句話說，它要麼增加1或0。第二個最右邊的數字呢？它要麼增加0或2。下一個？ 0或4.看到圖案？

讓我們寫的僞代碼：

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 

因爲我們開始與電力和共0，你就會明白爲什麼這個工程。

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)