我们如何比较两个特里的相似性？

问题描述

我只是好奇是否有一种方法可以比较两次try数据结构的相似性？

trie1                      trie2

   root                     root 
/     |                   /   |
m     b                   m   b
|     |                   |   |
a     o                   a   o
| \   |                   |   |
t  x  b                   x   b

def compare_trie(trie1,trie2):
    pass

Output["max","bob"]

编辑：到目前为止，我尝试实现dfs算法，但对如何管理两个不同尝试的堆栈感到震惊

我仍然尝试通过管理两次不同尝试的两个堆栈来敲击代码：

def compareTrie(trie1,trie2):
    dfsstack = []
    result = []
    stack1 = [x for x in trie1.keys()]
    stack2 = [y for y in trie2.keys()]
    similar = list(set(stack1) & set(stack2))
    dfsstack.append((similar,result))
    while (dfsstack):
        current,result = dfsstack.pop()
        print(current,result)
        result.append(current)
        for c in current:
            trie1 = trie1[c]
            trie2 = trie2[c]
            st1 = [x for x in trie1.keys()]
            st2 = [x for x in trie2.keys()]
            simm = list(set(st1) & set(st2))
            dfsstack.append((simm,result))

    print(result)

Trie实现：

def create_trie(words):
    trie = {}
    for word in words:
        curr = trie
        for c in word:
            if c not in curr:
                curr[c] = {}
            curr = curr[c]
        # Mark the end of a word
        curr['#'] = True
    return trie


s1 = "mat max bob"
s2 = "max bob"

words1 = s1.split()
words2 = s2.split()

t1 = create_trie(words1)
t2 = create_trie(words2)

解决方法

您使用dfs的想法是正确的；但是，您可以选择一种简单的解决方法来解决当前的任务。这是递归版本：

def create_trie(words):
    trie = {}
    for word in words:
        curr = trie
        for c in word:
            if c not in curr:
                curr[c] = {}
            curr = curr[c]
        # Mark the end of a word
        curr['#'] = True
    return trie

def compare(trie1,trie2,curr):
    for i in trie1.keys():
        if trie2.get(i,None):
            if i=="#":
                result.append(curr)
            else:
                compare(trie1[i],trie2[i],curr+i)
    

s1 = "mat max bob temp2 fg f r"
s2 = "max bob temp fg r c"

words1 = s1.split()
words2 = s2.split()

t1 = create_trie(words1)
t2 = create_trie(words2)
result = []
compare(t1,t2,"")
print(result)   #['max','bob','fg','r']

,

对于当前状态，可以用一个堆栈替换递归。并在result方法内创建compare数组。

def compare(trie1,trie2):
    result = []
    stack = [(trie1,"")]
    while stack:
        t1,curr = stack.pop()
        for i in t1:
            if i not in t2:
                continue
            if i == "#":
                result.append(curr)
            else:
                stack.append((t1[i],t2[i],curr + i))
    return result

,

是的，有可能。大多数情况是，因为您在仅是图灵机不足的设备上使用通用语言。

执行此操作的微不足道的方法是遍历每个trie，生成一组所有键。拿这两个集合的交集。