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word_break.py
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word_break.py
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"""
139. 单词拆分
字典树 记忆化搜索 哈希表 字符串 动态规划
中等
给定一个非空字符串 s 和一个包含非空单词的列表 wordDict,判定 s 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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/word-break
"""
from typing import List
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
n = len(s)
dp = [False] * (n+1)
word_set = set(wordDict)
dp[0] = True
for i in range(n):
for j in range(i+1, n+1):
if dp[i] and s[i:j] in word_set:
dp[j] = True
return dp[-1]
if __name__ == '__main__':
solution = Solution()
result = solution.wordBreak("leetcode", ["leet", "code"])
print(result)
assert result is True
result = solution.wordBreak("applepenapple", ["apple", "pen"])
print(result)
assert result is True
result = solution.wordBreak("catsandog", ["cats", "dog", "sand", "and", "cat"])
print(result)
assert result is False