@@ -337,10 +337,53 @@ class Solution {
337337
338338Python:
339339
340+ 回溯
341+ ``` python
342+ class Solution :
343+ def backtracking (self s : str , wordSet : set[str ], startIndex : int ) -> bool :
344+ # 边界情况:已经遍历到字符串末尾,返回True
345+ if startIndex >= len (s):
346+ return True
347+ 348+ # 遍历所有可能的拆分位置
349+ for i in range (startIndex, len (s)):
350+ word = s[startIndex:i + 1 ] # 截取子串
351+ if word in wordSet and self .backtracking(s, wordSet, i + 1 ):
352+ # 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
353+ return True
354+ 355+ # 无法进行有效拆分,返回False
356+ return False
357+ 358+ def wordBreak (self s : str , wordDict : List[str ]) -> bool :
359+ wordSet = set (wordDict) # 转换为哈希集合,提高查找效率
360+ return self .backtracking(s, wordSet, 0 )
361+ 362+ ```
363+ DP(版本一)
364+ ``` python
365+ class Solution :
366+ def wordBreak (self s : str , wordDict : List[str ]) -> bool :
367+ wordSet = set (wordDict)
368+ n = len (s)
369+ 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
370+ dp[0 ] = True # 初始状态,空字符串可以被拆分成单词
371+ 372+ for i in range (1 , n + 1 ):
373+ for j in range (i):
374+ if dp[j] and s[j:i] in wordSet:
375+ 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
376+ break
377+ 378+ return dp[n]
379+ 380+ 381+ ```
382+ DP(版本二)
383+ 340384``` python
341385class Solution :
342386 def wordBreak (self s : str , wordDict : List[str ]) -> bool :
343- ''' 排列'''
344387 dp = [False ]* (len (s) + 1 )
345388 dp[0 ] = True
346389 # 遍历背包
@@ -351,17 +394,6 @@ class Solution:
351394 dp[j] = dp[j] or (dp[j - len (word)] and word == s[j - len (word):j])
352395 return dp[len (s)]
353396```
354- ``` python
355- class Solution : # 和视频中写法一致(和最上面C++写法一致)
356- def wordBreak (self s : str , wordDict : List[str ]) -> bool :
357- dp = [False ]* (len (s)+ 1 )
358- dp[0 ]= True
359- for j in range (1 ,len (s)+ 1 ):
360- for i in range (j):
361- word = s[i:j]
362- if word in wordDict and dp[i]: dp[j]= True
363- return dp[- 1 ]
364- ```
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