[leetcode]Compare Version Numbers

难度是Easy，开始没考虑那么多，直接撸。。。然后当然错了

简单的想法，把两个version split成两个num的list，然后再比较。。。

class Solution:
    # @param a, a string
    # @param b, a string
    # @return a boolean
 
    def compareVersion(self, version1, version2):
        def compareInt(a, b) :
            if (a > b):
                return 1
            elif (a < b):
                return -1
            else:
                return 0
        ver1 = version1.split('.')
        ver2 = version2.split('.')
        num1 = map(lambda x : int(x), ver1)
        num2 = map(lambda x : int(x), ver2)
        if (len(num1) < len(num2)) :
            for i in xrange(0, len(num2) - len(num1)):
                num1.append(0)
        if (len(num1) > len(num2)) :
            for i in xrange(0, len(num1) - len(num2)):
                num2.append(0)
        pos = filter(lambda (x, y) : x != y, zip(num1, num2))
        if (len(pos) != 0) :
            return compareInt(pos[0][0], pos[0][1])
        return compareInt(len(num1), len(num2))