Skip to content

Latest commit

 

History

History
105 lines (78 loc) · 2.85 KB

0648.md

File metadata and controls

105 lines (78 loc) · 2.85 KB

0648. 单词替换

  • 标签:字典树、数组、哈希、字符串
  • 难度:中等

题目链接

题目大意

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

输入dictionary = ["cat","bat","rat"], sentence = "the cattle was rattled by the battery"
输出"the cat was rat by the bat"


输入dictionary = ["a","b","c"], sentence = "aadsfasf absbs bbab cadsfafs"
输出"a a b c"

解题思路

思路 1:字典树

  1. 构造一棵字典树。
  2. 将所有的词根存入到前缀树(字典树)中。
  3. 然后在树上查找每个单词的最短词根。

思路 1:代码

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)

思路 1:复杂度分析

  • 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:$O(|dictionary| + |sentence|)$。