lintcode 中等题: Implement Trie

题目

Implement a trie with insert, search, and startsWith methods.

样例
 
注意

You may assume that all inputs are consist of lowercase letters a-z.

解题

    Trie,字典树,又称单词查找树、前缀树,是一种哈希树的变种。应用于字符串的统计与排序,经常被搜索引擎系统用于文本词频统计。

    性质:

    1.根节点不包含字符,除根节点外的每一个节点都只包含一个字符。

    2.从根节点到某一节点,路径上经过的字符连接起来,为该节点对应的字符串。

    3.每个节点的所有子节点包含的字符都不相同。

    优点是查询快。对于长度为m的键值,最坏情况下只需花费O(m)的时间;而BST需要O(m log n)的时间。

程序来源链接

1.理解Trie 字典树很重要

2.定义TrieNode节点类很重要

class TrieNode {
    // Initialize your data structure here.
    char c;  
    boolean leaf;  
    HashMap<Character, TrieNode> children = new HashMap<Character, TrieNode>();  
    public TrieNode(char c) {
        this.c = c;  
    }
    public TrieNode() {}
}

孩子节点是HashMap的形式,可以很快速的取出其中的值。

上面理解了下面插入删除就容易了

/**
 * Your Trie object will be instantiated and called as such:
 * Trie trie = new Trie();
 * trie.insert("lintcode");
 * trie.search("lint"); will return false
 * trie.startsWith("lint"); will return true
 */
class TrieNode {
    // Initialize your data structure here.
    char c;  
    boolean leaf;  
    HashMap<Character, TrieNode> children = new HashMap<Character, TrieNode>();  
    public TrieNode(char c) {
        this.c = c;  
    }
    public TrieNode() {}
}

public class Solution {
    private TrieNode root = null;

    public Solution() {
        root = new TrieNode();
    }

    // Inserts a word into the trie.
    public void insert(String word) {
        Map<Character,TrieNode> children = root.children;
        for(int i = 0;i< word.length() ;i++){
            char c = word.charAt(i);
            TrieNode t = null;
            if(children.containsKey(c)){
                t = children.get(c);
            }else{
                t = new TrieNode(c);
                children.put(c,t);
            }
            children = t.children;
            if(i == word.length() - 1)
                t.leaf = true;
        }
        
    }

    // Returns if the word is in the trie.
    public boolean search(String word) {
        TrieNode t = searchNode(word);
        return t!=null && t.leaf;
    }

    // Returns if there is any word in the trie
    // that starts with the given prefix.
    public boolean startsWith(String prefix) {
        return searchNode(prefix) != null;
    }
    
    private TrieNode searchNode(String word){
        Map<Character ,TrieNode> children = root.children;
        TrieNode t = null;
        for(int i = 0;i< word.length() ;i++){
            char c = word.charAt(i);
            if(!children.containsKey(c)) 
                return null;
            t = children.get(c);
            children = t.children;
        }
        return t;
    }
}
Java Code

总耗时: 1599 ms

原文地址:https://www.cnblogs.com/bbbblog/p/4943575.html