Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.
Note that the same word in the dictionary may be reused multiple times in the segmentation.
DP:
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): # 遍历背包
for j in range(i): # 遍历单词
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]backtracking:
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)
