总结一下Word Break I 和 II
第一题：
Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
For example, given
s = "leetcode",
dict = ["leet", "code"].
"leetcode" can be segmented as
"leet code"
第二题：
Given a string s and a dictionary of words dict, add spaces in s 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