Refactored algorithm for Challenge 13

challenge_13/python/alexbotello/README.md

@@ -2,22 +2,18 @@
 ###### Language Version (Python 3.6.0)
 
 
-We have our function take the integer input and convert it into a list of
-integers. We find the number of digits, the midpoint, and then separate both
-halves of the list.
+First, we initialize a int variable ```reverse``` and set it equal to zero. As
+well as make a copy variable of the integer input ```num2```. Using a while
+loop, our algorithm will continue until ```num2``` is equal to zero. 
 
-If the number of digits is even, reverse the second half and compare it to the
-first. If number of digits are odd, follow the same steps, but be sure to add +1
-to the index of the first half.
-
-Run ```palin.py`` to test your implementation
+Run ```palin.py``` to test your implementation
 ```
 456070654
 
 True
 ```
 
-Run ```tests.py`` to check against unit test
+Run ```tests.py``` to check against unit test
 ```
 .......
 ----------------------------------------------------------------------

challenge_13/python/alexbotello/src/palin.py

@@ -4,20 +4,18 @@ def is_palindrome(num):
     """
     Returns True if integer is a Palindrome
     """
-    num = [int(x) for x in str(num)]
-    digits = len(num)
-    mid = digits // 2
-    back_half = num[mid:]
+    reverse = 0
+    num2 = num
 
-    if digits % 2 == 0:
-        return True if num[:mid] == back_half[::-1] else False
-
-    else:
-        return True if num[:mid+1] == back_half[::-1] else False
+    while num2 > 0:
+        # Finds the last digit of num2, removes it, and adds the
+        # digit to the reverse variable
+        reverse = (10*reverse) + num2%10
+        num2 //= 10
 
+    return True if reverse == num else False
 
 if __name__ == "__main__":
-
     num = int(input())
     print()
     print(is_palindrome(num))