lintcode 中等题： Implement Trie

题目

Implement Trie

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