题目大意
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描述：给定一个由许多词根组成的字典列表 dictionary，以及一个句子字符串 sentence。
要求：将句子中有词根的单词用词根替换掉。如果单词有很多词根，则用最短的词根替换掉他。最后输出替换之后的句子。
说明：
- $1 \le dictionary.length \le 1000$。
- $1 \le dictionary[i].length \le 100$。
dictionary[i] 仅由小写字母组成。- $1 \le sentence.length \le 10^6$。
sentence 仅由小写字母和空格组成。sentence 中单词的总量在范围 $[1, 1000]$ 内。sentence 中每个单词的长度在范围 $[1, 1000]$ 内。sentence 中单词之间由一个空格隔开。sentence 没有前导或尾随空格。
示例：
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输入：dictionary = ["cat","bat","rat"], sentence = "the cattle was rattled by the battery"
输出："the cat was rat by the bat"
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输入：dictionary = ["a","b","c"], sentence = "aadsfasf absbs bbab cadsfafs"
输出："a a b c"
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解题思路
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思路 1：字典树
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- 构造一棵字典树。
- 将所有的词根存入到前缀树（字典树）中。
- 然后在树上查找每个单词的最短词根。
思路 1：代码
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class Trie:
def __init__(self):
"""
Initialize your data structure here.
"""
self.children = dict()
self.isEnd = False
def insert(self, word: str) -> None:
"""
Inserts a word into the trie.
"""
cur = self
for ch in word:
if ch not in cur.children:
cur.children[ch] = Trie()
cur = cur.children[ch]
cur.isEnd = True
def search(self, word: str) -> str:
"""
Returns if the word is in the trie.
"""
cur = self
index = 0
for ch in word:
if ch not in cur.children:
return word
cur = cur.children[ch]
index += 1
if cur.isEnd:
break
return word[:index]
class Solution:
def replaceWords(self, dictionary: List[str], sentence: str) -> str:
trie_tree = Trie()
for word in dictionary:
trie_tree.insert(word)
words = sentence.split(" ")
size = len(words)
for i in range(size):
word = words[i]
words[i] = trie_tree.search(word)
return ' '.join(words)
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思路 1：复杂度分析
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- 时间复杂度：$O(|dictionary| + |sentence|)$。其中 $|dictionary|$ 是字符串数组
dictionary 中的字符总数，$|sentence|$ 是字符串 sentence 的字符总数。
- 空间复杂度：$O(|dictionary| + |sentence|)$。