我们如何比较两个尝试的相似性？

Question

我只是好奇是否有办法比较两个尝试数据结构的相似性？

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，但对如何管理不同尝试的两个堆栈感到震惊

我尝试过的代码仍然通过管理两个堆栈进行两次不同的尝试而感到震惊：

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)

尝试实现：

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)

Answer 1

您使用 dfs 的想法是正确的；但是，您可以选择一种简单的递归方法来解决手头的任务。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']