Leetcode 139. Word Break
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
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
class Solution:
def wordBreak(self, s: str, wordDict: List) -> bool:
dic = set(wordDict)
hashmap = {}
return self.helper(s, dic, hashmap)
def helper(self, s: str, dic: List, hashmap) -> bool:
if s in hashmap:
return hashmap
if s in dic:
hashmap = True
return True
for i in range(1, len(s)):
left = s[:i]
right = s
if self.helper(left, dic, hashmap) and right in dic:
hashmap = True
return True
hashmap = False
return False
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