648. Replace Words (Medium)

Posted 2020-10-07 Yancea --- TO BE A DOER !

In English, we have a concept called root, which can be followed by some other words to form another longer word - let‘s call this word successor. For example, the root an, followed by other, which can form another word another.

Now, given a dictionary consisting of many roots and a sentence. You need to replace all the successor in the sentence with the rootforming it. If a successor has many roots can form it, replace it with the root with the shortest length.

You need to output the sentence after the replacement.

Example 1:

Input: dict = ["cat", "bat", "rat"]
sentence = "the cattle was rattled by the battery"
Output: "the cat was rat by the bat"

Note:

1. The input will only have lower-case letters.
2. 1 <= dict words number <= 1000
3. 1 <= sentence words number <= 1000
4. 1 <= root length <= 100
5. 1 <= sentence words length <= 1000

思路：字典树（Trie）
首先构建字典树，进行遍历，然后搜索；

class TrieNode():
    def __init__(self):
        self.data = {}
        self.is_word = False

class Trie():
    def __init__(self):
        self.root = TrieNode()
        
    def insert(self, word):
        node = self.root
        for letter in word:
            child = node.data.get(letter)
            if not child:
                node.data[letter] = TrieNode()
            node = node.data[letter]
        node.is_word = True
        
    def search(self, word):
        string = ‘‘
        node = self.root
        for letter in word:
            node = node.data.get(letter)
            if not node:
                break
            string += letter
            if node.is_word:
                return string
        return word
    
class Solution():
    def replaceWords(self, dict, sentence):
        """
        :type dict: List[str]
        :type sentence: str
        :rtype: str
        """
        trie = Trie()
        for word in dict: 
            trie.insert(word)
        res = []
        for word in sentence.split():
            res.append(trie.search(word))
        return ‘ ‘.join(res)