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Day 68

题目描述

思路

根据昨天的trie前缀树进行修改，特殊需要考虑的点在于存在通配符，我来说明下如何解决这个问题的：
关键在于这段代码

 for (WordDictionary child : words.child) {if (child != null && find(child, word, i + 1)) {return true;}}return false;

遍历当前节点的所有非空子节点，对每个子节点递归调用 find 函数，处理剩余字符（start + 1）。
只要找到一条有效路径，立即返回 true。
如果所有子节点都无法匹配，返回 false。
做法

class WordDictionary {public WordDictionary[]child;public boolean isend;public WordDictionary() {child=new WordDictionary[27];isend=false;}public void addWord(String word) {WordDictionary words=this;for(int i=0;i<word.length();i++){char x=word.charAt(i);int index=x-'a';if(words.child[index]==null){words.child[index]=new WordDictionary();}words=words.child[index];}words.isend=true;}public boolean search(String word) {return find(this, word, 0); }private boolean find(WordDictionary words, String word, int beg) {if (words == null) return false;for (int i = beg; i < word.length(); i++) {char c = word.charAt(i);if (c == '.') {for (WordDictionary child : words.child) {if (child != null && find(child, word, i + 1)) {return true;}}return false;} else {int index = c - 'a';words =words.child[index];if (words == null) return false;}}return words.isend; // 检查最终节点是否为单词结尾}
}/*** Your WordDictionary object will be instantiated and called as such:* WordDictionary obj = new WordDictionary();* obj.addWord(word);* boolean param_2 = obj.search(word);*/