标签：reset map cto lse trie har pytho python ace
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:
a-z.实现一个神奇字典，包含buildDict和search函数。buildDict函数的功能是能把给的没有重复单词的列表建立一个字典，search函数的功能是存在和这个单词只有一个位置上的字符不同返回true，否则返回false。
Java:
class MagicDictionary {
Map<String, List<int[]>> map = new HashMap<>();
/** Initialize your data structure here. */
public MagicDictionary() {
}
/** Build a dictionary through a list of words */
public void buildDict(String[] dict) {
for (String s : dict) {
for (int i = 0; i < s.length(); i++) {
String key = s.substring(0, i) + s.substring(i + 1);
int[] pair = new int[] {i, s.charAt(i)};
List<int[]> val = map.getOrDefault(key, new ArrayList<int[]>());
val.add(pair);
map.put(key, val);
}
}
}
/** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
public boolean search(String word) {
for (int i = 0; i < word.length(); i++) {
String key = word.substring(0, i) + word.substring(i + 1);
if (map.containsKey(key)) {
for (int[] pair : map.get(key)) {
if (pair[0] == i && pair[1] != word.charAt(i)) return true;
}
}
}
return false;
}
}
Python:
class MagicDictionary(object):
def _candidates(self, word):
for i in xrange(len(word)):
yield word[:i] + ‘*‘ + word[i+1:]
def buildDict(self, words):
self.words = set(words)
self.near = collections.Counter(cand for word in words
for cand in self._candidates(word))
def search(self, word):
return any(self.near[cand] > 1 or
self.near[cand] == 1 and word not in self.words
for cand in self._candidates(word))
Python:
# Time: O(n), n is the length of the word
# Space: O(d)
import collections
class MagicDictionary(object):
def __init__(self):
"""
Initialize your data structure here.
"""
_trie = lambda: collections.defaultdict(_trie)
self.trie = _trie()
def buildDict(self, dictionary):
"""
Build a dictionary through a list of words
:type dictionary: List[str]
:rtype: void
"""
for word in dictionary:
reduce(dict.__getitem__, word, self.trie).setdefault("_end")
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
"""
def find(word, curr, i, mistakeAllowed):
if i == len(word):
return "_end" in curr and not mistakeAllowed
if word[i] not in curr:
return any(find(word, curr[c], i+1, False) for c in curr if c != "_end") if mistakeAllowed else False
if mistakeAllowed:
return find(word, curr[word[i]], i+1, True) or any(find(word, curr[c], i+1, False) for c in curr if c not in ("_end", word[i]))
return find(word, curr[word[i]], i+1, False)
return find(word, self.trie, 0, True)
C++:
class MagicDictionary {
public:
/** Initialize your data structure here. */
MagicDictionary() {}
/** Build a dictionary through a list of words */
void buildDict(vector<string> dict) {
for (string word : dict) {
m[word.size()].push_back(word);
}
}
/** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
bool search(string word) {
for (string str : m[word.size()]) {
int cnt = 0, i = 0;
for (; i < word.size(); ++i) {
if (word[i] == str[i]) continue;
if (word[i] != str[i] && cnt == 1) break;
++cnt;
}
if (i == word.size() && cnt == 1) return true;
}
return false;
}
private:
unordered_map<int, vector<string>> m;
};
C++:
class MagicDictionary {
public:
/** Initialize your data structure here. */
MagicDictionary() {}
/** Build a dictionary through a list of words */
void buildDict(vector<string> dict) {
for (string word : dict) s.insert(word);
}
/** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
bool search(string word) {
for (int i = 0; i < word.size(); ++i) {
char t = word[i];
for (char c = ‘a‘; c <= ‘z‘; ++c) {
if (c == t) continue;
word[i] = c;
if (s.count(word)) return true;
}
word[i] = t;
}
return false;
}
private:
unordered_set<string> s;
};
类似题目：
[LeetCode] 208. Implement Trie (Prefix Tree) 实现字典树(前缀树)
720. Longest Word in Dictionary
[LeetCode] 676. Implement Magic Dictionary 实现神奇字典
标签：reset map cto lse trie har pytho python ace
原文地址：https://www.cnblogs.com/lightwindy/p/9808248.html
