8.23 - hard - 100

527. Word Abbreviation

基本的想法就是先计算出所有的abbr，然后把相同的abbr的index记录下来，然后依次扩展直到这一位消除了所有的相同的，这个video简直是神解释：https://www.youtube.com/watch?v=yAQMcGY4c90

class Solution(object):
    def wordsAbbreviation(self, dict):
        """
        :type dict: List[str]
        :rtype: List[str]
        """
        def get_abbreviation(word,old_abbreviation):
            if old_abbreviation == '':
                abbreviation = word[0] + str(len(word[1:-1])) + word[-1]                
            else:
                for i in range(len(old_abbreviation)):
                    if old_abbreviation[i].isdigit():
                        abbreviation = word[:i+1] + str(len(word[i+1:-1])) + word[-1]
                        break
            if len(word) <= len(abbreviation):
                abbreviation = word
            return abbreviation
        
        
        abbreviations = [''] * len(words)
        duplicates = collections.defaultdict(list)
        for i,word in enumerate(words):
            abbreviations[i] = get_abbreviation(word,'')
            duplicates[abbreviations[i]] += i,
        
        for i in range(len(abbreviations)):
            if len(duplicates[abbreviations[i]]) == 1:
                continue
            else:
                while len(duplicates[abbreviations[i]]) > 1:
                    for index in duplicates[abbreviations[i]]:
                        abbreviation = get_abbreviation(words[index], abbreviations[index])
                        abbreviations[index] = abbreviation
                        duplicates[abbreviation] += index,
                        
        return abbreviations