总结一下Word Break I 和 II

第一题：

Given a string s and a dictionary of words dict, determine ifs can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code"

第二题：

Given a string s and a dictionary of words dict, add spaces ins to construct a sentence where each word is a valid dictionary word.

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"].

第一题是判断字典里是否存在S的可行的拆分，第二题是求所有可行的拆分。动态规划。第二题需要先利用第一题的结果来判断是否存在拆分，再计算拆分，否则会超时。

class Solution(object):    def wordBreak(self, s, wordDict):        """        :type s: str        :type wordDict: Set[str]        :rtype: bool        """        partial = [False] * len(s)        for i in range(len(s)):            if s[:i+1] in wordDict:                partial[i] = True                continue            for j in range(i-1, -1, -1):                if partial[j] == True and s[j+1:i+1] in wordDict:                    partial[i] = True                    break        return partial[len(s) - 1]                    

class Solution(object):    def wordBreak(self, s, wordDict):        """        :type s: str        :type wordDict: Set[str]        :rtype: List[str]        """        divisible = [False] * len(s)        for i in range(len(s)):            if s[:i+1] in wordDict:                divisible[i] = True                continue            for j in range(0, i):                if divisible[j] and s[j+1:i+1] in wordDict:                    divisible[i] = True                    break                results = []        if not divisible[len(s) - 1]:            return results                    candidates = [[] for i in range(len(s))]        for i in range(len(s)):            cur = s[:i+1]            if cur in wordDict:                candidates[i].append([cur])            for j in range(0, i):                new = s[j+1:i+1]                if new in wordDict and len(candidates[j]) != 0:                    for k in candidates[j]:                        candidates[i].append(k + [new])                                for i in candidates[len(s) - 1]:            results.append(" ".join(i))                    return results