边工作边刷题：70天一遍leetcode: day 93

Generalized Abbreviation

要点：不算难题，注意这类递归的题要按super set的方式（选or不选），而不是combination。比较容易理解。

• 这题的限制条件是abbreviation的部分要连成一片。
  • 如果是subset方法，因为是从左向右，所以记录左边的count，然后要不reset，然后填入字母，或者继续count+=1。
  • 如果是combination的方法，每选k（k>=0）个进入下一层。注意要在本层把下一个字母加上，下层要从该字母之后开始，这样不会出现连续数字的情况
• 这里涉及一个何时结束的问题。这类题的要点都是在下一个位置更新，因为最后是结束在index==n的位置
• 注意res用+=创建新object比较好，不用考虑backtrack的问题。反正最后也要copy之后添加到solutions里

https://repl.it/CbpH/1

# Write a function to generate the generalized abbreviations of a word.

# Example:
# Given word = "word", return the following list (order does not matter):
# ["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]


class Solution(object):
    def generateAbbreviations(self, word):
        """
        :type word: str
        :rtype: List[str]
        """
        def helper(word, res, solutions):
            if not word:
                solutions.append(res)
                return
            
            helper(word[1:], res+word[0], solutions)
            
            for i in xrange(len(word)):
                helper(word[i+2:], res+str(i+1)+word[i+1:i+2], solutions)
        
        solutions = []
        helper(word, "", solutions)
        return solutions

sol = Solution()
assert sol.generateAbbreviations('word')==["word","wor1","wo1d","wo2","w1rd","w1r1","w2d","w3","1ord","1or1","1o1d","1o2","2rd","2r1","3d","4"]