今天學習到了Trie樹
字典樹的詳細定義可以看
Stackoverflow上的簡潔實踐
How to create a trie in Python
class Trie:
def __init__(self):
self._end = '_end_'
def make_trie(self,*words):
root = dict()
for word in words:
current_dict = root
for letter in word:
current_dict = current_dict.setdefault(letter,{})
current_dict[self._end] = self._end
return root
if __name__ == '__main__':
trie = Trie()
print(trie.make_trie('foo', 'bar', 'baz', 'barz'))
控制檯輸出
{'f': {'o': {'o': {'_end_': '_end_'}}},
'b': {'a': {'r': {'_end_': '_end_', 'z': {'_end_': '_end_'}}, 'z': {'_end_': '_end_'}}}}
在trie樹中查找單詞
class Trie:
def __init__(self):
self._end = '_end_'
def make_trie(self,*words):
root = dict()
for word in words:
current_dict = root
for letter in word:
current_dict = current_dict.setdefault(letter,{})
current_dict[self._end] = self._end
return root
def in_trie(self,trie,word):
current_dict = trie
for letter in word:
if letter not in current_dict:
return False
current_dict = current_dict[letter]
return self._end in current_dict
if __name__ == '__main__':
trie = Trie()
print(trie.make_trie('foo', 'bar', 'baz', 'barz'))
print(trie.in_trie(trie.make_trie('foo', 'bar', 'baz', 'barz'),'bar'))
# output: True
單詞的壓縮編碼(leetcode)
給定一個單詞列表,我們將這個列表編碼成一個索引字符串 S 與一個索引列表 A。
例如,如果這個列表是 [“time”, “me”, “bell”],我們就可以將其表示爲 S = “time#bell#” 和
indexes = [0, 2, 5]。對於每一個索引,我們可以通過從字符串 S 中索引的位置開始讀取字符串,直到 “#” 結束,來恢復我們之前的單詞列表。
那麼成功對給定單詞列表進行編碼的最小字符串長度是多少呢?
示例:
輸入: words = [“time”, “me”, “bell”] 輸出: 10 說明: S = “time#bell#” ,
indexes = [0, 2, 5] 。來源:力扣(LeetCode)
鏈接:https://leetcode-cn.com/problems/short-encoding-of-words
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我覺得leetcode的官方實現纔是最6的,pythonic極致
class Solution:
def minimumLengthEncoding(self, words: List[str]) -> int:
words = list(set(words)) #去重
# Trie是帶有已創建節點的嵌套字典
# 當其中缺少節點時會創建節點
Trie = lambda: collections.defaultdict(Trie)
trie = Trie()
#reduce(..., S, trie) is trie[S[0]][S[1]][S[2]][...][S[S.length - 1]],將單詞反序插入
nodes = [reduce(dict.__getitem__, word[::-1], trie)
for word in words]
#如果節點沒有鄰居節點,則添加該單詞
return sum(len(word) + 1
for i, word in enumerate(words)
if len(nodes[i]) == 0)
作者:LeetCode-Solution
鏈接:https://leetcode-cn.com/problems/short-encoding-of-words/solution/dan-ci-de-ya-suo-bian-ma-by-leetcode-solution/
來源:力扣(LeetCode)
著作權歸作者所有。商業轉載請聯繫作者獲得授權,非商業轉載請註明出處。
奇怪的知識又增加了哈哈哈~
今早還學習到了
set.discard(ele)# 可以移除集合中不存在的元素