648. 单词替换

2021/04/04 leetcode 共 1352 字,约 4 分钟

title

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在英语中,我们有一个叫做 词根(root)的概念,它可以跟着其他一些词组成另一个较长的单词——我们称这个词为 继承词(successor)。例如,词根an,跟随着单词 other(其他),可以形成新的单词 another(另一个)。

现在,给定一个由许多词根组成的词典和一个句子。你需要将句子中的所有继承词用词根替换掉。如果继承词有许多可以形成它的词根,则用最短的词根替换它。

你需要输出替换之后的句子。

示例 1:

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

示例 2:

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

示例 3:

输入:dictionary = ["a", "aa", "aaa", "aaaa"], sentence = "a aa a aaaa aaa aaa aaa aaaaaa bbb baba ababa"
输出:"a a a a a a a a bbb baba a"

示例 4:

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

示例 5:

输入:dictionary = ["ac","ab"], sentence = "it is abnormal that this solution is accepted"
输出:"it is ab that this solution is ac"

提示:

  • 1 <= dictionary.length <= 1000
  • 1 <= dictionary[i].length <= 100
  • dictionary[i] 仅由小写字母组成。
  • 1 <= sentence.length <= 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。

solution

class Solution:
    def replaceWords(self, dictionary: List[str], sentence: str) -> str:
        trie = {}
        for word in dictionary:
            node = trie
            for c in word:
                if c not in node:
                    node[c] = {}
                node = node[c]
            node['#'] = None

        def search(word):
            node = trie
            for i, c in enumerate(word):
                if c not in node:
                    return word
                node = node[c]
                if "#" in node:
                    return word[:i+1]
            return word
        words = sentence.split(" ")
        ans = []
        for word in words:
            ans.append(search(word))
        return " ".join(ans)

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