Check Word Abbreviation

这题算法就是模拟，实现没有任何难度，唯一麻烦就是各种各样奇诡的boundary case导致很难一次性bug free搞定，这里只是把比较好懂的一段代码放上来，碰到也只能背板了如果想bug free的话

class Solution(object):
    def validWordAbbreviation(self, word, abbr):
        """
        :type word: str
        :type abbr: str
        :rtype: bool
        """
        cur_abrr = 0
        cur_word = 0
        n = len(word)
        m = len(abbr)

        while cur_abrr < m:
            if abbr[cur_abrr].isalpha():
                if abbr[cur_abrr] != word[cur_word]:                        # 直接字符对应不相等
                    return False
                cur_abrr += 1
                cur_word += 1
            elif abbr[cur_abrr] == '0':                                     # 缩写串里面有先导0不符合题意
                return False
            else:
                tmp = 0
                while cur_abrr < m and abbr[cur_abrr].isdigit():
                    tmp = 10 * tmp + ord(abbr[cur_abrr]) - ord('0')
                    cur_abrr += 1
                cur_word += tmp
                if cur_word > n or (cur_word == n and cur_abrr < m):        # 考虑原字串已经超过终点值以及原字串到达终点但缩写串没有
                    return False

        return True if cur_word == n and cur_abrr == m else False           # 最后以防万一再check一遍是不是同时到达了终点

放一种相对简洁一点的版本，这里主要是先导零的问题很烦，思路上还是one edit distance的改版

class Solution(object):
    def validWordAbbreviation(self, word, abbr):
        """
        :type word: str
        :type abbr: str
        :rtype: bool
        """

        wordPt, abbrPt = 0, 0
        n, m = len(word), len(abbr)
        curNum = 0

        while wordPt < n and abbrPt < m:
            if abbr[abbrPt].isalpha():
                wordPt += curNum
                curNum = 0
                if wordPt >= n or abbr[abbrPt] != word[wordPt]:
                    return False
                wordPt += 1
            else:
                curNum = 10 * curNum + ord(abbr[abbrPt]) - ord('0')
                if not curNum:
                    return False
            abbrPt += 1

        return wordPt + curNum == n and abbrPt == m