Fraction to Recurring Decimal/ Excel Sheet Column Number

Author: gavinfish

166 Fraction to Recurring Decimal.py

@@ -0,0 +1,44 @@
+'''
+Given two integers representing the numerator and denominator of a fraction, return the fraction in string format.
+
+If the fractional part is repeating, enclose the repeating part in parentheses.
+
+For example,
+
+Given numerator = 1, denominator = 2, return "0.5".
+Given numerator = 2, denominator = 1, return "2".
+Given numerator = 2, denominator = 3, return "0.(6)".
+Hint:
+
+No scary math, just apply elementary math knowledge. Still remember how to perform a long division?
+Try a long division on 4/9, the repeating part is obvious. Now try 4/333. Do you see a pattern?
+Be wary of edge cases! List out as many test cases as you can think of and test your code thoroughly.
+'''
+
+class Solution(object):
+    def fractionToDecimal(self, numerator, denominator):
+        """
+        :type numerator: int
+        :type denominator: int
+        :rtype: str
+        """
+        sign = '-' if numerator * denominator < 0 else ''
+        quotient, remainder = divmod(abs(numerator), abs(denominator))
+        result_list = [sign, str(quotient), '.']
+        remainders = []
+        while remainder not in remainders:
+            remainders.append(remainder)
+            quotient, remainder = divmod(remainder * 10, abs(denominator))
+            result_list.append(str(quotient))
+
+        idx = remainders.index(remainder)
+        result_list.insert(idx + 3, '(')
+        result_list.append(')')
+        result = ''.join(result_list).replace('(0)', '').rstrip('.')
+        return result
+
+
+if __name__ == "__main__":
+    assert Solution().fractionToDecimal(1, 2) == '0.5'
+    assert Solution().fractionToDecimal(2, 1) == '2'
+    assert Solution().fractionToDecimal(2, 3) == '0.(6)'

171 Excel Sheet Column Number.py

@@ -0,0 +1,33 @@
+'''
+Related to question Excel Sheet Column Title
+
+Given a column title as appear in an Excel sheet, return its corresponding column number.
+
+For example:
+
+    A -> 1
+    B -> 2
+    C -> 3
+    ...
+    Z -> 26
+    AA -> 27
+    AB -> 28 
+'''
+
+class Solution(object):
+    def titleToNumber(self, s):
+        """
+        :type s: str
+        :rtype: int
+        """
+        base = ord('A') - 1
+        n = len(s)
+        result = 0
+        for i in range(n):
+            result += (ord(s[n - 1 - i]) - base) * pow(26, i)
+        return result
+
+
+if __name__ == "__main__":
+    assert Solution().titleToNumber('A') == 1
+    assert Solution().titleToNumber('AB') == 28