Update 0139.单词拆分.md

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jianghongcheng
2023-06-05 03:09:10 -05:00
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parent 45037cec33
commit d8c51b2d36

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@ -337,10 +337,53 @@ class Solution {
Python
回溯
```python
class Solution:
def backtracking(self, s: str, wordSet: set[str], startIndex: int) -> bool:
# 边界情况已经遍历到字符串末尾返回True
if startIndex >= len(s):
return True
# 遍历所有可能的拆分位置
for i in range(startIndex, len(s)):
word = s[startIndex:i + 1] # 截取子串
if word in wordSet and self.backtracking(s, wordSet, i + 1):
# 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
return True
# 无法进行有效拆分返回False
return False
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
wordSet = set(wordDict) # 转换为哈希集合,提高查找效率
return self.backtracking(s, wordSet, 0)
```
DP版本一
```python
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
wordSet = set(wordDict)
n = len(s)
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
dp[0] = True # 初始状态,空字符串可以被拆分成单词
for i in range(1, n + 1):
for j in range(i):
if dp[j] and s[j:i] in wordSet:
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
break
return dp[n]
```
DP版本二
```python
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
'''排列'''
dp = [False]*(len(s) + 1)
dp[0] = True
# 遍历背包
@ -351,17 +394,6 @@ class Solution:
dp[j] = dp[j] or (dp[j - len(word)] and word == s[j - len(word):j])
return dp[len(s)]
```
```python
class Solution: # 和视频中写法一致和最上面C++写法一致)
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
dp = [False]*(len(s)+1)
dp[0]=True
for j in range(1,len(s)+1):
for i in range(j):
word = s[i:j]
if word in wordDict and dp[i]: dp[j]=True
return dp[-1]
```