|
马上注册,结交更多好友,享用更多功能^_^
您需要 登录 才可以下载或查看,没有账号?立即注册
x
- Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
- Note:
- The same word in the dictionary may be reused multiple times in the segmentation.
- You may assume the dictionary does not contain duplicate words.
- 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
复制代码
- 给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict,判定是否可以被空格拆分为一个或多个在字典中出现的单词。
- 说明:
- 拆分时可以重复使用字典中的单词。
- 你可以假设字典中没有重复的单词。
- 示例 1:
- 输入: s = "leetcode", wordDict = ["leet", "code"]
- 输出: true
- 解释: 返回 true 因为 "leetcode" 可以被拆分成 "leet code"。
- 示例 2:
- 输入: s = "applepenapple", wordDict = ["apple", "pen"]
- 输出: true
- 解释: 返回 true 因为 "applepenapple" 可以被拆分成 "apple pen apple"。
- 注意你可以重复使用字典中的单词。
- 示例 3:
- 输入: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
- 输出: false
复制代码
- class Solution:
- def wordBreak(self, s: str, wordDict: List[str]) -> bool:
- dic = set(wordDict)
- hashmap = {}
- return self.helper(s, dic, hashmap)
-
- def helper(self, s: str, dic: List[str], hashmap) -> bool:
- if s in hashmap:
- return hashmap[s]
- if s in dic:
- hashmap[s] = True
- return True
- for i in range(1, len(s)):
- left = s[:i]
- right = s[i:]
- if self.helper(left, dic, hashmap) and right in dic:
- hashmap[s] = True
- return True
- hashmap[s] = False
- return False
复制代码 |
|