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Update 0139.单词拆分.md
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@ -337,10 +337,53 @@ class Solution {
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Python:
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回溯
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```python
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class Solution:
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def backtracking(self, s: str, wordSet: set[str], startIndex: int) -> bool:
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# 边界情况:已经遍历到字符串末尾,返回True
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if startIndex >= len(s):
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return True
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# 遍历所有可能的拆分位置
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for i in range(startIndex, len(s)):
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word = s[startIndex:i + 1] # 截取子串
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if word in wordSet and self.backtracking(s, wordSet, i + 1):
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# 如果截取的子串在字典中,并且后续部分也可以被拆分成单词,返回True
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return True
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# 无法进行有效拆分,返回False
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return False
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def wordBreak(self, s: str, wordDict: List[str]) -> bool:
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wordSet = set(wordDict) # 转换为哈希集合,提高查找效率
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return self.backtracking(s, wordSet, 0)
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```
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DP(版本一)
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```python
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class Solution:
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def wordBreak(self, s: str, wordDict: List[str]) -> bool:
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wordSet = set(wordDict)
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n = len(s)
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dp = [False] * (n + 1) # dp[i] 表示字符串的前 i 个字符是否可以被拆分成单词
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dp[0] = True # 初始状态,空字符串可以被拆分成单词
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for i in range(1, n + 1):
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for j in range(i):
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if dp[j] and s[j:i] in wordSet:
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dp[i] = True # 如果 s[0:j] 可以被拆分成单词,并且 s[j:i] 在单词集合中存在,则 s[0:i] 可以被拆分成单词
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break
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return dp[n]
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```
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DP(版本二)
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```python
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class Solution:
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def wordBreak(self, s: str, wordDict: List[str]) -> bool:
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'''排列'''
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dp = [False]*(len(s) + 1)
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dp[0] = True
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# 遍历背包
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@ -351,17 +394,6 @@ class Solution:
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dp[j] = dp[j] or (dp[j - len(word)] and word == s[j - len(word):j])
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return dp[len(s)]
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```
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```python
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class Solution: # 和视频中写法一致(和最上面C++写法一致)
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def wordBreak(self, s: str, wordDict: List[str]) -> bool:
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dp = [False]*(len(s)+1)
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dp[0]=True
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for j in range(1,len(s)+1):
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for i in range(j):
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word = s[i:j]
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if word in wordDict and dp[i]: dp[j]=True
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return dp[-1]
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```
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