Valid Word Abbreviation

字符串的处理实在太脏了，有时候不是写不出来，而是五花八门的corner case使得逻辑无法干净，因为无法一次性考虑这么多的边界情况，不过双指针比对的题，现在看起来最外层循环条件还是把两个指针的判断都放进去最好

class Solution(object):
    def validWordAbbreviation(self, word, abbr):
        """
        :type word: str
        :type abbr: str
        :rtype: bool
        """
        n, m = len(word), len(abbr)
        curWord, curAbbr = 0, 0
        while curWord < n and curAbbr < m:
            if word[curWord] == abbr[curAbbr]:
                curWord += 1
                curAbbr += 1
                continue
            if abbr[curAbbr] == '0' or not abbr[curAbbr].isdigit():
                return False
            curVal = 0
            while curAbbr < m and abbr[curAbbr].isdigit():
                curVal = 10 * curVal + ord(abbr[curAbbr]) - ord('0')
                curAbbr += 1
            curWord += curVal
        return curWord == n and curAbbr == m