传递要放入特里的字符串列表

问题描述

当给定一个字符串时，我有可以构建特里数据结构的代码。当我尝试传递字符串列表时，它将单词组合成一个

class TrieNode:
    def __init__(self):
        self.end = False
        self.children = {}

    def all_words(self,prefix):
        if self.end:
            yield prefix

        for letter,child in self.children.items():
            yield from child.all_words(prefix + letter)

class Trie:
    def __init__(self):
        self.root = TrieNode()
    def __init__(self):
        self.root = TrieNode()

    def insert(self,words):
        curr = self.root
        #the line I added to read the words from a list is below
        for word in words:
            for letter in word:
                node = curr.children.get(letter)
                if not node:
                    node = TrieNode()
                    curr.children[letter] = node
                curr = node
            curr.end  = True


    def all_words_beginning_with_prefix(self,prefix):
        cur = self.root
        for c in prefix:
            cur = cur.children.get(c)
            if cur is None:
                return  # No words with given prefix

        yield from cur.all_words(prefix)

这是我用来将所有内容插入树中的代码：

lst = ['foo','foob','foobar','foof']
trie = Trie()
trie.insert(lst)

我得到的输出是

['foo','foofoob','foofoobfoobar','foofoobfoobarfoof']

我想要的输出是

['foo','foof']

这是我用来获取输出的行（为了重现性，以防您需要运行代码）-它返回所有以特定前缀开头的单词：

print(list(trie.all_words_beginning_with_prefix('foo')))

我该如何解决？

解决方法

每次插入后，您无需将curr重置为根，因此您要在上一个中断的位置插入下一个单词。您想要类似的东西：

def insert(self,words):
    curr = self.root
    for word in words:
        for letter in word:
            node = curr.children.get(letter)
            if not node:
                node = TrieNode()
                curr.children[letter] = node
            curr = node
        curr.end  = True
        curr = self.root  # Reset back to the root

不过我会分解的。我认为您的insert函数做得太多，并且不应该处理多个字符串。我将其更改为：

def insert(self,word):
    curr = self.root
    for letter in word:
        node = curr.children.get(letter)
        if not node:
            node = TrieNode()
            curr.children[letter] = node
        curr = node
    curr.end  = True

def insert_many(self,words):
    for word in words:
        self.insert(word)  # Just loop over self.insert

现在这不是问题，因为每个insert是一个独立的调用，您不能忘记重置curr。