标签：匹配 append 查询系统 als sig continue i++ ESS return
题意： 设计个字典查询系统， 有 add 和search 两种操作， add 是加入单词到字典里， search 时 可以用 点号通配符 "."， 点号可以匹配一个字母。
分析： 当search 时为 通配符时， 如果直接用back tracking产生 a-z, 比如 有7个点号， 就得生成 26^7 个组合，会TLE。
以下是TLE的code:
class WordDictionary { /** Initialize your data structure here. */ Set<String> dic; public WordDictionary() { dic = new HashSet<>(); } /** Adds a word into the data structure. */ public void addWord(String word) { dic.add(word); } /** Returns if the word is in the data structure. A word could contain the dot character ‘.‘ to represent any one letter. */ public boolean search(String word) { return dfs(new StringBuilder(), word, 0); } private boolean dfs(StringBuilder curResult, String word, int index){ if(curResult.length() == word.length()){ // System.out.println(curResult.toString()); if(dic.contains(curResult.toString())) return true; return false; } boolean success = false; char cur_ch = word.charAt(index); if(cur_ch == ‘.‘){ for(int i=0; i<26; i++){ curResult.append((char)(i+‘a‘)); success = success || dfs(curResult,word,index+1); curResult.setLength(curResult.length()-1); if(success) return true; } } else { curResult.append(cur_ch); success = success || dfs(curResult,word,index+1); curResult.setLength(curResult.length()-1); if(success) return true; } return success; } } /** * Your WordDictionary object will be instantiated and called as such: * WordDictionary obj = new WordDictionary(); * obj.addWord(word); * boolean param_2 = obj.search(word); */
改进: 用map 来存放 <长度+ List<String> > 的组合， 匹配一个单词 首先得长度匹配，才能进一步匹配。
换成如下 算法能beat 99%, 但如果单词长度全部一样， 那 变成了 n^2的算法了。主要还是测试数据太弱了。
class WordDictionary { /** Initialize your data structure here. */ Map<Integer, List<String>> dict; public WordDictionary() { dict = new HashMap<>(); } /** Adds a word into the data structure. */ public void addWord(String word) { List<String> val = dict.getOrDefault(word.length(),new ArrayList<>()) ; val.add(word); dict.put(word.length(), val); } /** Returns if the word is in the data structure. A word could contain the dot character ‘.‘ to represent any one letter. */ public boolean search(String word) { if(!dict.containsKey(word.length())) return false; List<String> list = dict.get(word.length()); int i; for(String str: list){ for(i=0; i<word.length(); i++){ char c = word.charAt(i); if(c == ‘.‘) continue; if(c != str.charAt(i)) break; } if(i == word.length()) return true; } return false; } }
Trie 的算法 待续。
211 Add and Search Word - Data structure design--- back tracking, map, set 待续 trie
标签：匹配 append 查询系统 als sig continue i++ ESS return
原文地址：https://www.cnblogs.com/keepAC/p/9972770.html
