Add Strings – Leetcode Challenge – Python Solution
This is the python solution for the Leetcode problem – Add Strings – Leetcode Challenge – Python Solution.
Source – qiyuangong’s repository.
class Solution(object):
# def addStrings(self, num1, num2):
# """
# :type num1: str
# :type num2: str
# :rtype: str
# """
# if num1 is None:
# num1 = '0'
# if num2 is None:
# num2 = '0'
# res = []
# carry = 0
# ls = min(len(num1), len(num2))
# pos = -1
# while pos + ls >= 0:
# curr = int(num1[pos]) + int(num2[pos]) + carry
# res.insert(0, str(curr % 10))
# carry = curr / 10
# pos -= 1
# while pos + len(num1) >= 0:
# curr = int(num1[pos]) + carry
# res.insert(0, str(curr % 10))
# carry = curr / 10
# pos -= 1
# while pos + len(num2) >= 0:
# curr = int(num2[pos]) + carry
# res.insert(0, str(curr % 10))
# carry = curr / 10
# pos -= 1
# if carry != 0:
# res.insert(0, str(carry))
# return ''.join(res)
def addStrings(self, num1, num2):
res = []
pos1 = len(num1) - 1
pos2 = len(num2) - 1
carry = 0
# This conditon is great
# https://leetcode.com/problems/add-strings/discuss/90436/Straightforward-Java-8-main-lines-25ms
while pos1 >= 0 or pos2 >= 0 or carry == 1:
digit1 = digit2 = 0
if pos1 >= 0:
digit1 = ord(num1[pos1]) - ord('0')
if pos2 >= 0:
digit2 = ord(num2[pos2]) - ord('0')
res.append(str((digit1 + digit2 + carry) % 10))
carry = (digit1 + digit2 + carry) / 10
pos1 -= 1
pos2 -= 1
# reverse res
return ''.join(res[::-1])
