使用OPsplit
方法及实施Trie
在发现Java 中的 Trie 数据结构 https://www.baeldung.com/trie-javaBaeldung的文章,我能够得到以下结果:
realways=real ways
arealways=a real ways
但是,如果我从字典中删除单词“real”或“a”,我会得到以下结果:
realways=null
arealways=are always
这是我用来获得这些结果的完整代码:
public class Splitter {
private static Map<String, String> map = new HashMap<>();
private Trie dict;
public Splitter(Trie t) {
dict = t;
}
/**
* @param args
*/
public static void main(String[] args) {
List<String> words = List.of("a", "always", "are", "area", "r", "way", "ways"); // The order of these words does not seem to impact the final result
String test = "arealways";
Trie t = new Trie();
for (String word : words) {
t.insert(word);
}
System.out.println(t);
Splitter splitter = new Splitter(t);
splitter.split(test);
map.entrySet().forEach(System.out::println);
}
public String split(String test) {
if (dict.find(test)) {
return (test);
} else if (map.containsKey(test)) {
return (map.get(test));
} else {
for (int i = 0; i < test.length(); i++) {
String pre = test.substring(0, i);
if (dict.find(pre)) {
String end = test.substring(i);
String fixedEnd = split(end);
if (fixedEnd != null) {
map.put(test, pre + " " + fixedEnd);
return pre + " " + fixedEnd;
} else {
}
}
}
}
map.put(test, null);
return null;
}
public static class Trie {
private TrieNode root = new TrieNode();
public boolean find(String word) {
TrieNode current = root;
for (int i = 0; i < word.length(); i++) {
char ch = word.charAt(i);
TrieNode node = current.getChildren().get(ch);
if (node == null) {
return false;
}
current = node;
}
return current.isEndOfWord();
}
public void insert(String word) {
TrieNode current = root;
for (char l : word.toCharArray()) {
current = current.getChildren().computeIfAbsent(l, c -> new TrieNode());
}
current.setEndOfWord(true);
}
@Override
public String toString() {
return toString(root);
}
/**
* @param root2
* @return
*/
private String toString(TrieNode node) {
return node.toString();
}
public static class TrieNode {
private Map<Character, TrieNode> children = new HashMap<>() ;
private String contents;
private boolean endOfWord;
public Map<Character, TrieNode> getChildren() {
return children;
}
public void setEndOfWord(boolean endOfWord) {
this.endOfWord = endOfWord;
}
public boolean isEndOfWord() {
return endOfWord;
}
@Override
public String toString() {
StringBuilder sbuff = new StringBuilder();
if (isLeaf()) {
return sbuff.toString();
}
children.entrySet().forEach(entry -> {
sbuff.append(entry.getKey() + "\n");
});
sbuff.append(" ");
return children.toString();
}
private boolean isLeaf() {
return children.isEmpty();
}
}
public void delete(String word) {
delete(root, word, 0);
}
private boolean delete(TrieNode current, String word, int index) {
if (index == word.length()) {
if (!current.isEndOfWord()) {
return false;
}
current.setEndOfWord(false);
return current.getChildren().isEmpty();
}
char ch = word.charAt(index);
TrieNode node = current.getChildren().get(ch);
if (node == null) {
return false;
}
boolean shouldDeleteCurrentNode = delete(node, word, index + 1) && !node.isEndOfWord();
if (shouldDeleteCurrentNode) {
current.getChildren().remove(ch);
return current.getChildren().isEmpty();
}
return false;
}
}
}
我通过添加一个改进了原始代码toString()
方法到Trie
and TrieNode
。现在,当我打印出Trie
对象“t”,我得到以下结果:
{a={r={e={a=}}, l={w={a={y={s=}}}}}, w={a={y={s=}}}}
我的结论是OP的TrieNode
实施不正确。方式Trie
已构建,给定输入的字符串值,OP 描述的行为似乎是正确的。