Word Break II

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.

Return all such possible sentences.

For example, given
s = “catsanddog”,
dict = [“cat”, “cats”, “and”, “sand”, “dog”].

A solution is [“cats and dog”, “cat sand dog”].

题目：其实就是给定一个字符串S，求解出该字符串能分解成给定字典单词组成的单词串的所有可能解
以下是leetcode discuss上大神stefan的python解

def wordBreak(self, s, wordDict):
    memo = {len(s): ['']}
    #当s为空串''时，len(s)=0，直接返回['']，十分优雅
    #
    def sentences(i):
        if i not in memo:
            memo[i] = [s[i:j] + (tail and ' ' + tail)
                       for j in range(i+1, len(s)+1)
                       if s[i:j] in wordDict
                       for tail in sentences(j)]
        return memo[i]
    return sentences(0)

该算法实际上是一个递规过程。