- Trie字典树/前缀树的直观感受
- Trie只用来处理字符串
其中蓝色就是单词结尾节点。
class Node{
boolean isWord; //表示该字母是否是单词的结尾
Map<char,Node> next;
}public class Trie {
private class Node{
public boolean isWord;//标记该字符是否是单词结尾
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;
}
//获取Trie中存储的单词数量
public int getSize(){
return size;
}
//向Trie中添加一个新单词word
public void add(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);
}
//循环结束后，cur不一定是叶子节点，比如Trie中已经有 "panda"，此时add("pan"),
// cur指向'n'节点，显然'n'不是叶子节点,那么就要标记为结束位置
if(!cur.isWord){
//!cur.isWord 表示该节点未被标识为结束位置
cur.isWord=true;
size++;
}
}
}//查询单词是否在Trie中
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);
}
//注意：即使循环结束了，也不一定能确定该单词就在Trie中
//如果Trie中已经有单词"panda"，此时要查询"pan"
//循环结束后,cur此时指向'n'节点,'n'节点不是结尾节点,即"pan"不在Trie中
return cur.isWord;
}//查询是否在Trie中存在以prefix为前缀的单词
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);
}
//注意：循环结束后,cur不管是单词的结尾节点还是非结尾节点，都成立
//单词本身就是该单词的前缀
return true;
}- LeetCode 208题 实现Trie字典树
class Trie {
private class Node{
public boolean isWord;//标记该字符是否是单词结尾
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 Trie() {
root=new Node();
}
/** Inserts a word into the trie. */
public void insert(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);
}
//循环结束后，cur不一定是叶子节点，比如Trie中已经有 "panda"，此时add("pan"),
// cur指向'n'节点，显然'n'不是叶子节点,那么就要标记为结束位置
if(!cur.isWord){
//!cur.isWord 表示该节点未被标识为结束位置
cur.isWord=true;
}
}
/** Returns if the word is in the trie. */
public boolean search(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);
}
//注意：即使循环结束了，也不一定能确定该单词就在Trie中
//如果Trie中已经有单词"panda"，此时要查询"pan"
//循环结束后,cur此时指向'n'节点,'n'节点不是结尾节点,即"pan"不在Trie中
return cur.isWord;
}
/** Returns if there is any word in the trie that starts with the given prefix. */
public boolean startsWith(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);
}
//注意：循环结束后,cur不管是单词的结尾节点还是非结尾节点，都成立
//单词本身就是该单词的前缀
return true;
}
}LeetCode 211
class WordDictionary {
private class Node{
public boolean isWord;//标记该字符是否是单词结尾
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);
}
//循环结束后，cur不一定是叶子节点，比如Trie中已经有 "panda"，此时add("pan"),
// cur指向'n'节点，显然'n'不是叶子节点,那么就要标记为结束位置
if(!cur.isWord){
//!cur.isWord 表示该节点未被标识为结束位置
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);
}
//判断word在index位置是否匹配
private boolean match(Node node,String word,int index){
if(index==word.length()){
return node.isWord;
}
char c=word.charAt(index);
if(c!='.') {
//c是小写字母
if (node.next.get(c) == null) {
return false;
}
return match(node.next.get(c), word, index + 1);
}else{
//遍历所有从以该点为根节点的子树
for(char nextChar:node.next.keySet()){
if(match(node.next.get(nextChar),word,index+1)){
return true;
}
}
return false;
}
}
}LeetCode 677
class MapSum {
private class Node{
public Node[] next;
public int value;
public Node(int value){
this.next=new Node[26];
this.value=value;
}
public Node(){
this(0);
}
}
private Node root;
/** Initialize your data structure here. */
public MapSum() {
root=new Node();
}
public void insert(String key, int val) {
Node cur=root;
for(int i=0;i<key.length();i++){
char c=key.charAt(i);
if(cur.next[c-'a']==null){
cur.next[c-'a']=new Node();
}
cur=cur.next[c-'a'];
}
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[c-'a']==null){
return 0;
}
cur=cur.next[c-'a'];
}
return sum(cur);
}
private int sum(Node node){
//说明node是叶子结点
if(node.next==null){
return node.value;
}
int res=node.value;
//node.next 是当前结点的所有的后继结点
for(Node nextNode:node.next){
if(nextNode!=null){
res+=sum(nextNode);
}
}
return res;
}
}