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给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict,在字符串中增加空格来构建一个句子,使得句子中所有的单词都在词典中。返回所有这些可能的句子。
说明:
分隔时可以重复使用字典中的单词。
你可以假设字典中没有重复的单词。
示例 1:
输入:
s = “catsanddog”
wordDict = [“cat”, “cats”, “and”, “sand”, “dog”]
输出:
[
“cats and dog”,
“cat sand dog”
]
示例 2:
输入:
s = “pineapplepenapple”
wordDict = [“apple”, “pen”, “applepen”, “pine”, “pineapple”]
输出:
[
“pine apple pen apple”,
“pineapple pen apple”,
“pine applepen apple”
]
解释: 注意你可以重复使用字典中的单词。
示例 3:
输入:
s = “catsandog”
wordDict = [“cats”, “dog”, “sand”, “and”, “cat”]
输出:
[]
先判断能否拆分,再dfs得出结果:
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
# 先判断能否拆分,再回溯生成所有结果
dp = [False] * (1+len(s))
dp[0], self.res = True, []
for i in range(1, len(s)+1):
for j in range(i):
if dp[j] and s[j:i] in wordDict:
dp[i] = True
break
if dp[-1]:
self.dfs(s, wordDict, '')
return self.res
def dfs(self, s, words, tmp):
if not s:
self.res.append(tmp[:-1])
for word in words:
if s.startswith(word):
self.dfs(s[len(word):], words, tmp+word+' ')
用dp的结果简化运算:
class Solution:
def wordBreak(self, s: str, wordDict):
# 先判断能否拆分,再回溯生成所有结果
self.dp = [False] * (1 + len(s))
self.dp[0], self.res = True, []
for i in range(1, len(s) + 1):
for j in range(i):
if self.dp[j] and s[j:i] in wordDict:
self.dp[i] = True
break
if self.dp[-1]:
self.dfs(s, wordDict, '', 0)
return self.res
def dfs(self, s, words, tmp, i):
if i == len(s):
self.res.append(tmp[:-1])
for j in range(1, len(s[i:])+1):
if self.dp[i+j] and s[i:i+j] in words:
self.dfs(s, words, tmp+s[i:i+j]+' ', i+j)