648. 单词替换

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。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/replace-words
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

 

class Solution:
    def replaceWords(self, dictionary: List[str], sentence: str) -> str:
        sentence=list(sentence.split())
        res=[]
        for i in sentence:
            swap=[]
            for j in dictionary:
                if i.startswith(j):
                    swap.append(j)
            if not swap:
                res.append(i) 
            else:
                res.append(min(swap))
        return ' '.join(res)

 

class Solution(object):
    def replaceWords(self, roots, sentence):
        """
        :type dict: List[str]
        :type sentence: str
        :rtype: str
        """
        # Because the default constructor registered by defaultdict is only called when it is called for the first time, so you can define yourself here.
        # Trie is a function, calling it will return a defaultdict
        Trie = lambda:collections.defaultdict(Trie)
        # tri first calls Trie, returns a defalutdict, so tri is a defalutdict
        tri = Trie()
        END = True
        
        for root in roots:
            # dict.__getitem__ requires two parameters, the first is the dictionary object, the second is the key
            # At the beginning, the dictionary object is tri, the key is the first character c of root, and tri[c] will return a new dictionary.
            # New dictionary represents the child node of the node starting with the character c
            # By analogy, finally get a leaf node, it should be empty by default, we assign the value of the whole word to the place where the key is END
            reduce(dict.__getitem__, root, tri)[END] = root
        
        def replace(word):
            cur = tri
            for c in word:
                # Either this character is not in the successor node of the current node, or has encountered the shortest prefix
                if c not in cur or END in cur:break
                # Otherwise continue to traverse down
                cur = cur[c]
            # When returning, there are two possibilities. If cur does not have a corresponding prefix, then the word does not match the word, and the word is returned.
            # get operation does not affect the value of Trie
            return cur.get(END, word)