https://leetcode.com/problems/add-binary/#/description
Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100"
.
Solution:
We will use recursion to solve this problem. First, we get last digit of the two numbers, a and b, and then if the last digits are 1 and 1, then we need to carry. We will call the addBinary method on the digits before the last digit, and then append 0 to the sum of two binary numbers. However, if the last digits are 0,0 or 0, 1, then we do not to carry. And we will recursively call the method on the previous digits and append 0 or 1 to the last digit of the sum.
class Solution(object): def addBinary(self, a, b): """ :type a: str :type b: str :rtype: str """ if len(a) == 0: return b if len(b) == 0: return a if a[-1] == '1' and b[-1] == '1': return self.addBinary(self.addBinary(a[:-1], b[:-1]), '1') + '0' if a[-1] == '0' and b[-1] == '0': return self.addBinary(a[:-1], b[:-1]) + '0' else: return self.addBinary(a[:-1], b[:-1]) + '1'
Note :
1. a[:-1] means get everything before the last element. Supper useful in digit retrival, espeacially the digits of number are unknown. Forget about the mod function, ugh.
ex.
a = [1,2,3,4,5]
a[:-1]
[1,2,3,4]
2. Edge check.
3. def in the class defines a METHOD not a FUNCTION. Do use the self.nameofmethod(...) when calling the method recursively.
ex.
class Solution:
def addBinary(self, a, b):
XXXXXXX
XXXXXXX
XXXXXXX self.addBinary(a,b) # no need to include self in the parentheses.
XXXXXXX
4. Python syntax:
if XXXXXXX:
XXXXXXX
if XXXXXXX:
XXXXXXX
else: # not elif
XXXXXXX
5. Use single quotes to quote numbers/strings.....
It is okay to add '1'/'0' at the end of or in the front of numbers, on bit manipulation.