## 题目描述：

LeetCode 676. Implement Magic Dictionary

Implement a magic directory with buildDict, and search methods.

For the method buildDict, you'll be given a list of non-repetitive words to build a dictionary.

For the method search, you'll be given a word, and judge whether if you modify exactly one character into another character in this word, the modified word is in the dictionary you just built.

Example 1:

Input: buildDict(["hello", "leetcode"]), Output: Null
Input: search("hello"), Output: False
Input: search("hhllo"), Output: True
Input: search("hell"), Output: False
Input: search("leetcoded"), Output: False

Note:

1. You may assume that all the inputs are consist of lowercase letters a-z.
2. For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
3. Please remember to RESET your class variables declared in class MagicDictionary, as static/class variables are persisted across multiple test cases. Please see here for more details.

## 题目大意：

buildDict：输入一组无重复的单词，构建一个字典

search：在字典中寻找目标串，目标串与搜索串恰好有一个字符不同

## 解题思路：

build操作：用下划线'_'替换word的每一个位置的字母，作为Key，被替换的字母作为Value，存入dmap

search操作：用下划线'_'替换word的每一个位置的字母，作为Key，在dmap中查找与被替换字母不同的值

## Python代码：

class MagicDictionary(object):

def __init__(self):
"""
"""
self.dmap = collections.defaultdict(set)

def buildDict(self, dict):
"""
Build a dictionary through a list of words
:type dict: List[str]
:rtype: void
"""
for word in dict:
for x in range(len(word)):
key = word[:x] + '_' + word[x+1:]

def search(self, word):
"""
Returns if there is any word in the trie that equals to the given word after modifying exactly one character
:type word: str
:rtype: bool
"""
for x in range(len(word)):
key = word[:x] + '_' + word[x+1:]
values = self.dmap[key]
if not values: continue
if word[x] not in values or len(values) > 1:
return True
return False

# Your MagicDictionary object will be instantiated and called as such:
# obj = MagicDictionary()
# obj.buildDict(dict)
# param_2 = obj.search(word)

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