[leetcode] 139. Word Break
阿新 • • 發佈:2022-04-06
題目
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.
Example 1:
Input: s = "leetcode", wordDict = ["leet","code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code".
Example 2:
Input: s = "applepenapple", wordDict = ["apple","pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
Note that you are allowed to reuse a dictionary word.
Example 3:
Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"] Output: false
Constraints:
1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
-
s
andwordDict[i]
consist of only lowercase English letters. - All the strings of
wordDict
are unique.
思路
動態規劃
程式碼
python版本:
# dfs,超時 class Solution: def wordBreak(self, s: str, wordDict: List[str]) -> bool: words = Counter(wordDict) def dfs(now): if now == len(s): return True nexts = [] for i in range(now+1, len(s)+1): if s[now:i] in words: nexts.append(i) return any([dfs(i) for i in nexts]) return dfs(0) # dp class Solution: def wordBreak(self, s: str, wordDict: List[str]) -> bool: dp = [False]*len(s) for i in range(len(s)): for word in wordDict: if i-len(word)+1 >= 0 and s[i-len(word)+1:i+1] == word and (i-len(word) < 0 or dp[i-len(word)]): dp[i] = True return dp[-1]