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Reverse And Add/신재�

ReverseAndAdd/신재�

~cpp 
import unittest

class ReverseAndAdder:
    def reverse(self, number):
        numberStr = str(number)
        reverseStr = ''
        for i in range(len(numberStr)):
            reverseStr += numberStr[len(numberStr) - i - 1]
        return int(reverseStr)

    def isPalindrome(self, number):
        numberStr = str(number)
        for i in range(len(numberStr)):
            if(numberStr[i] != numberStr[len(numberStr) - i - 1]):
                return False
        return True

    def reverseAndAdd(self, number):
        count = 0
        for i in range(1000):
            number = number + self.reverse(number)
            count += 1
            if(self.isPalindrome(number)):
                break
        else:
            print 'No reverse and add'
            return None
        return (count, number)
        
class ReverseAndAdderTestCase(unittest.TestCase):
    def testReverse(self):
        raa = ReverseAndAdder()
        self.assertEquals(1234, raa.reverse(4321))
        self.assertEquals(12321, raa.reverse(12321))

    def testIsPalindrome(self):
        raa = ReverseAndAdder()
        self.assertEquals(False, raa.isPalindrome(4321))
        self.assertEquals(True, raa.isPalindrome(12321))
        self.assertEquals(True, raa.isPalindrome(1221))

    def testReverseAndAdd(self):
        raa = ReverseAndAdder()
        self.assertEquals((4, 9339), raa.reverseAndAdd(195))
        self.assertEquals((5, 45254), raa.reverseAndAdd(265))
        self.assertEquals((3, 6666), raa.reverseAndAdd(750))
        
if __name__ == '__main__':
    #unittest.main()
    raa = ReverseAndAdder()
    result = raa.reverseAndAdd(int(input()))
    if(result != None):
        print str(result[0]) + ' ' + str(result[1])

그리 맘� 들지는 않는다. --재�

reverse 메소드 대신� 다�과 같� 간단히 할 수 있다:

~cpp 
>>> n=4321
>>> int(str(n)[::-1]) #python 2.3��
1234
>>> int(''.join(reversed(str(d)))) #python 2.4��
1234

파���서 for 루프를 먼저 ��하는 것� 사고� 단위가 작� 것�고, �것� �문�해를 할 때 한번� 한단어씩 보는 것과 구문� 한번� 보는 것� 차�와 비�할 수� 있다.

위 코드를 �용하면...

~cpp 
>>> def rev(d):	return int(str(d)[::-1])

>>> def revAndAdd(d):
	c=0
	while True:
		rd=rev(d)
		if d==rd:break
		d+=rd
		c+=1
	return c,d

>>> revAndAdd(195)
(4, 9339)

all tests data will be computable with less than 1000 iterations (additions)를 고려한다면 명시�� 회수 체�는 없어� � 듯.

--JuNe

마치 pseudo code를 보는 것 같네요. --재�
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