我试图在Java中实现一个非常简单的Trie，支持3个操作。我希望它有一个insert方法，一个has方法（即trie中的某个单词），以及一个以字符串形式返回trie的toString方法。我相信我的插入工作正常，但是并且toString被证明是困难的。这是我到目前为止所拥有的。
特里班。
public class CaseInsensitiveTrie implements SimpleTrie {
//root node
private TrieNode r;
public CaseInsensitiveTrie() {
r = new TrieNode();
}
public boolean has(String word) throws InvalidArgumentUosException {
return r.has(word);
}
public void insert(String word) throws InvalidArgumentUosException {
r.insert(word);
}
public String toString() {
return r.toString();
}
public static void main(String[] args) {
CaseInsensitiveTrie t = new CaseInsensitiveTrie();
System.out.println("Testing some strings");
t.insert("TEST");
t.insert("TATTER");
System.out.println(t.has("TEST"));
}
}
和节点类
public class TrieNode {
//make child nodes
private TrieNode[] c;
//flag for end of word
private boolean flag = false;
public TrieNode() {
c = new TrieNode[26]; //1 for each letter in alphabet
}
protected void insert(String word) {
int val = word.charAt(0) - 64;
//if the value of the child node at val is null, make a new node
//there to represent the letter
if (c[val] == null) {
c[val] = new TrieNode();
}
//if word length > 1, then word is not finished being added.
//otherwise, set the flag to true so we know a word ends there.
if (word.length() > 1) {
c[val].insert(word.substring(1));
} else {
c[val].flag = true;
}
}
public boolean has(String word) {
int val = word.charAt(0) - 64;
if (c[val]!=null && word.length()>1) {
c[val].has(word.substring(1));
} else if (c[val].flag==true && word.length()==1) {
return true;
}
return false;
}
public String toString() {
return "";
}
}
所以基本上，在创建Trie时，TrieNode被创建为具有26个子节点的根。尝试插入时，将在该根节点上调用insert，该节点以正确的位置递归创建新节点，并继续直到该单词完成。我相信这种方法运作正常。
我的函数非常破碎，因为出于某种原因我必须在括号外面有返回语句。我不能在else子句中包含它或编译器抱怨。除此之外，我认为这种方法应该有一些调整，但我无法弄清楚我的生活。
toString是一个我试图解决的野兽，但是我没有投入任何东西，所以我会离开它，直到我解决了问题。如果我开始工作，我可能想办法将其重新格式化为toString函数。
int val = word.charAt（0） - 64的目的;是因为输入的每个字符串必须全部大写（我将创建一个字符串格式化函数以确保此后）所以第一个字母的int值 - 64将是它在数组中的位置。即，数组索引0是A，所以A = 64，A = 64 = 0. B = 65，B = 64 = 1，依此类推。
你的has函数应该看起来像这样：
if (c[val]!=null && word.length()>1) {
return c[val].has(word.substring(1)); //<-- Change is on this line
} else if (c[val].flag==true && word.length()==1) {
...etc
您执行递归调用，但您确实需要让该值传播回原始调用者。
也许您可以使用“Map c”而不是“TrieNode [] c”，这将允许您将此用于所有类型的字符大写/小写甚至特殊字符甚至可以节省空间（在26处分配26个字符数组）每个角色级别）
这是我的实施： -
public class Tries {
class Node {
HashMap<Character, Node> children;
boolean end;
public Node(boolean b){
children = new HashMap<Character, Tries.Node>();
end = false;
}
}
private Node root;
public Tries(){
root = new Node(false);
}
public static void main(String args[]){
Tries tr = new Tries();
tr.add("dog");
tr.add("doggy");
System.out.println(tr.search("dogg"));
System.out.println(tr.search("doggy"));
}
private boolean search(String word) {
Node crawl = root;
int n = word.length();
for(int i=0;i<n;i++){
char ch = word.charAt(i);
if(crawl.children.get(ch) == null){
return false;
}
else {
crawl = crawl.children.get(ch);
if(i==n-1 && crawl.end == true){
return true;
}
}
}
return false;
}
private void add(String word) {
Node crawl = root;
int n = word.length();
for(int i=0;i<n;i++){
char ch = word.charAt(i);
if(crawl.children.containsKey(ch)){
crawl = crawl.children.get(ch);
}
else {
crawl.children.put(ch, new Node(false));
Node temp = crawl.children.get(ch);
if(i == n-1){
temp.end = true;
}
crawl = temp;
System.out.println(ch + " " + crawl.end);
}
}
}
}
在这里我的实施：
public class Tries {
private static class Leaf {
private Leaf(char c) {
this.c=c;
}
char c;
int counter = 1;
List<Leaf> leaves = new ArrayList<>(10);
}
private Leaf root = new Leaf('0');
public void add(String word) {
Leaf current = root;
Leaf newLeaf = null;
for (char c : word.toCharArray()) {
boolean found = false;
for (Leaf leaf : current.leaves) {
if (leaf.c == c) {
current = leaf;
current.counter++;
found=true;
break;
}
}
if (!found) {
newLeaf = new Leaf(c);
current.leaves.add(newLeaf);
current = newLeaf;
}
}
}
public int find(String partial) {
Leaf current = root;
for (char c : partial.toCharArray()) {
boolean found = false;
for (Leaf leaf : current.leaves) {
if (leaf.c == c) {
current=leaf;
found=true;
break;
}
}
if(!found) return 0;
}
return current.counter;
}
public boolean hasWord(String partial) {
return find(partial)>0;
}
}
这是简单的java实现，不使用任何其他数据结构
import java.util.ArrayList;
import java.util.List;
class Trie {
private static Node root = new Node(' ', false);
static int getIndex(char x) {
return ((int) x) - ((int) 'a');
}
static class Node {
char data;
boolean isLeaf;
Node[] children;
Node(char data, boolean leaf) {
this.data = data;
this.isLeaf = leaf;
this.children = new Node[26];
}
}
static void insert(String data, Node root) {
if (data == null || data.length() == 0) {
return;
}
Node child = root.children[getIndex(data.charAt(0))];
if (child == null) {
Node node = new Node(data.charAt(0), data.length() == 1);
root.children[getIndex(data.charAt(0))] = node;
if (data.length() > 1) {
insert(data.substring(1, data.length()), node);
}
} else {
if (data.length() == 1) {
child.isLeaf = true;
} else {
insert(data.substring(1, data.length()), child);
}
}
}
static boolean find(String data, Node root) {
if (data == null || data.length() == 0) {
return true;
}
char x = data.charAt(0);
//note that first node ie root is just dummy, it just holds important
Node node = root.children[getIndex(x)];
if (node == null) {
return false;
} else {
if (data.length() == 1) {
return node.isLeaf;
} else {
return find(data.substring(1, data.length()), node);
}
}
}
static boolean delete(String data, Node root) {
if (data == null || data.length() == 0) {
return false;
}
char x = data.charAt(0);
//note that first node ie root is just dummy, it just holds important
Node node = root.children[getIndex(x)];
if (node == null) {
return false;
} else {
if (data.length() == 1) {
node.isLeaf = false;
boolean allNull = true;
for (Node node1 : node.children) {
allNull = allNull && node1 == null;
}
return allNull;
} else {
boolean delete = delete(data.substring(1, data.length()), node);
if (delete) {
node.children[getIndex(x)] = null;
if(node.isLeaf){
return false;
}
boolean allNull = true;
for (Node node1 : node.children) {
allNull = allNull && node1 == null;
}
return allNull; }
}
}
return false;
}
private static List<String> strings = new ArrayList<>();
private static List<String> getAll() {
strings = new ArrayList<String>();
findAllDFS(root, "");
return strings;
}
private static void findAllDFS(Node node, String old) {
if (node != null) {
if (node.data != ' ') {
old = old + node.data;
}
if (node.isLeaf) {
strings.add(old);
}
for (Node node1 : node.children) {
findAllDFS(node1, old);
}
}
}
public static void main(String[] args) {
insert("abc", root);
insert("xyz", root);
insert("abcd", root);
insert("abcde", root);
delete("abcd", root);
/* System.out.println(find("abc", root));
System.out.println(find("abcd", root));
System.out.println(find("ab", root));
System.out.println(find("xyz", root));*/
System.out.println(getAll());
}
}