LeetCode211. 添加与搜索单词 - 数据结构设计(相关话题:Trie前缀树)

LeetCode211. 添加与搜索单词 - 数据结构设计(相关话题:Trie前缀树)

 LeetCode211. 添加与搜索单词 - 数据结构设计(相关话题:Trie前缀树)

 LeetCode211. 添加与搜索单词 - 数据结构设计(相关话题:Trie前缀树)

import java.util.TreeMap;
class WordDictionary {
    private class Node{
        public boolean isWord;
        public TreeMap<Character, Node> next;

        public Node(boolean isWord){
            this.isWord = isWord;
            next = new TreeMap<>();
        }

        public Node(){
            this(false);
        }
    }
    private Node root;

    /** Initialize your data structure here. */
    public WordDictionary() {
        root = new Node();

    }
    
    /** Adds a word into the data structure. */
    public void addWord(String word) {
        Node cur = root;
        for(int i = 0; i < word.length(); i++){
            char c = word.charAt(i);
            if(cur.next.get(c) == null){ // 当前节点的next中是否存在字符c对应的下一个节点的映射
                cur.next.put(c, new Node());
            }
            cur = cur.next.get(c);
        }
        cur.isWord = true;

    }
    
    /** 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 match(root, word, 0);

    }
    private boolean match(Node node, String word, int index){
        if(index == word.length()){
            return node.isWord;
        }
        char c = word.charAt(index);
        if(c != ‘.‘){
            if(node.next.get(c) == null){
                return false;
            }
            return match(node.next.get(c), word, index + 1);
        }
        // 等于‘.‘的情况是node的下一个字符的所有可能都去匹配
        else{
            for(char nextChar : node.next.keySet()){
                if(match(node.next.get(nextChar), word, index + 1)){
                    return true;
                }
            }
            return false;

        }
    }
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * boolean param_2 = obj.search(word);
 */

Solution bobo

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