传递要放入 trie 中的字符串列表

Question

我有代码可以在给定一个字符串时构建一个 trie 数据结构。当我试图传递一个字符串列表时，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']

我想要得到的输出是

['foo', 'foob', 'foobar', 'foof']

这是我用来获取输出的行（为了可重复性，以防您需要运行代码）- 它 returns 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？

Answer 1

您不会在每次插入后将 curr 重置回词根，因此您是在上一个单词停止的地方插入下一个单词。你想要这样的东西：

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

虽然我会打破它。我认为您的 insert 函数做的太多了，不应该处理多个字符串。我会将其更改为：

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

现在这是一个 non-problem，因为每个 insert 都是一个独立的调用，您不能忘记重置 curr。

传递要放入 trie 中的字符串列表

Passing a list of strings to be put into trie

python

data-structures

trie