题目：原题链接（困难）
标签：动态规划、回溯算法、记忆化递归
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( 2 N ) 其中N为字符串长度 | O ( D ) : 其中D为字典词数 | 超出时间限制 |
Ans 2 (Python) | O ( 2 N ) | O ( D ) | 44ms (89%) |
Ans 3 (Python) |
解法一（回溯算法）：
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
if not wordDict:
return []
word_set = set(wordDict)
size = max(len(word) for word in word_set)
def dfs(s1):
ans = []
if s1 in word_set:
ans.append(s1)
for i in range(1, len(s1)):
if s1[:i] in word_set:
for s2 in dfs(s1[i:]):
ans.append(s1[:i] + " " + s2)
if i > size:
break
return ans
return dfs(s)解法二（记忆化递归）：
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
if not wordDict:
return []
word_set = set(wordDict)
size = max(len(word) for word in word_set)
@functools.lru_cache(None)
def dfs(s1):
ans = []
if s1 in word_set:
ans.append(s1)
for i in range(1, len(s1)):
if s1[:i] in word_set:
for s2 in dfs(s1[i:]):
ans.append(s1[:i] + " " + s2)
if i > size:
break
return ans
return dfs(s)
