一、什么是Trie
Trie不同于二分搜索树、堆、线段树等二叉树结构,Trie是一个多叉树。使用场景:通讯录高效搜索,专为处理字符串设计的。
比如字典中有n条数据,如果使用树结构,查询的时间复杂度是O(logn),如果有100万条数据的话,logn大约是20,如果有1亿条数据的话,logn大约是30(参考2的N次方计算器)
如果使用Trie这种数据结构,查询每条数据的时间复杂度和字典中一共有多少条数据没有关系!是不是屌炸天呢?
Trie查询的时间复杂度与查询的字符长度有关,时间复杂度为:O(w),w为单词的长度。
二、构建一个Trie
Trie的基本结构与添加方法:
public class Trie {
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;
private int size;
public Trie() {
root = new Node();
size = 0;
}
//获得Trie中存储的单词数量
public int getSize(){
return this.size;
}
//传递入一个字符串(单词),拆分成一个个的字符char
public void add(String word){
Node cur = root;
for(int i = 0 ; i < word.length() ; i ++){
char c = word.charAt(i);
//判断当前的cur节点下一节点隐射中是否有指向c的节点
if(cur.next.get(c) == null)
cur.next.put(c,new Node());
//循环结束后cur来到字符串最后一个字符所处节点,但并不一定是叶子节点,如pan和panda
cur = cur.next.get(c);
}
//如果已经存在panda,则在add(pan)时候,只是走了3遍cur = cur.next.get(c);
//不重复添加元素
if(!cur.isWord) {
cur.isWord = true;
size++;
}
}
}
判断某个单词在Trie中是否存在
public boolean contains(String word){
Node cur = root;
for(int i = 0 ; i < word.length() ; i++){
char c = word.charAt(i);
if(cur.next.get(c) == null)
return false;
cur = cur.next.get(c);
}
//循环结束后则表示到达了单词结尾的字符
return cur.isWord;
}
三、Trie字典树的前缀查询
//Trie字典树的前缀查询
public boolean isPrefix(String prefix){
Node cur = root;
for(int i = 0 ; i < prefix.length() ; i++){
char c = prefix.charAt(i);
if(cur.next.get(c) == null)
return false;
cur = cur.next.get(c);
}
return true;
}
四、Trie字典树搜索和正则匹配
对应Leetcode 211题添加与搜索单词
参考模型:
import java.util.TreeMap;
/**
* Your WordDictionary object will be instantiated and called as such:
* WordDictionary obj = new WordDictionary();
* obj.addWord(word);
* boolean param_2 = obj.search(word);
*/
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)
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);
}else{
for(char nextChar : node.next.keySet())
if(match(node.next.get(nextChar),word,index + 1))
return true;
return false;
}
}
}
五、Letcode键值映射——前缀开头的键的值的总和
对应Letcode 667题 键值映射
代码实现:
class MapSum {
private class Node{
private int value;
private TreeMap<Character,Node> next;
public Node(int value){
this.value = value;
next = new TreeMap<>();
}
public Node(){
this(0);
}
}
private Node root;
/** Initialize your data structure here. */
public MapSum() {
root = new Node();
}
public void insert(String word, int val) {
Node cur = root;
for(int i = 0 ; i < word.length() ; i++){
char c = word.charAt(i);
if(cur.next.get(c) == null)
cur.next.put(c,new Node());
cur = cur.next.get(c);
}
cur.value = val;
}
public int sum(String prefix) {
Node cur = root;
for(int i = 0 ; i < prefix.length() ; i++){
char c = prefix.charAt(i);
if(cur.next.get(c) == null)
return 0;
cur = cur.next.get(c);
}
return sum(cur);
}
private int sum(Node node) {
int res = node.value;
for(char c : node.next.keySet())
res += sum(node.next.get(c));
return res;
}
}