我正在尝试编写一个程序,该程序将接收来自dictionary.txt文档的输入并将这些单词存储在词典列表中,然后确定该单词是否可以分为三个或更多单词,如果可以,则进行打印原来的单词后面跟着新单词,例如disconsolateness:disc on so lateness
将是组成More.txt文件的输出。现在,代码只是继续运行,但是我没有得到任何输出,也不确定自己做错了什么。您能提供的任何帮助将不胜感激。我已经在下面发布了我的代码,输入的内容是字典中的任何单词。
import java.util.*;
import java.io.*;
public class CompositionTwo
{
private static List<String> dictionary = new ArrayList<>();
public static void main(String []args) {
File inputFile = new File("dictionary.txt");
File outputFile = new File("composedMore.txt");
Scanner in = null;
PrintWriter out = null;
try {
in = new Scanner(inputFile);
out = new PrintWriter(outputFile);
while (in.hasNext()) {
String input = in.next();
dictionary.add(input);
String output = splitWord(input, "");
if (output != "") {
out.println(input + ":" + output);
}
}
} catch (IOException e) {
System.out.println("An IOException has occurred during output process.");
} finally {
in.close();
out.close();
}
}
public static String splitWord(String word, String output) {
if (word.length() == 0) {
output = output;
}
else {
for (int i = 1; i <= word.length(); i++) {
// consider all prefixes of current String
String prefix = word.substring(0, i);
// if the prefix is present in the dictionary, add prefix to the
// output String and recurse for remaining String
if (dictionary.contains(prefix)) {
splitWord(word.substring(i), output + " " + prefix);
}
}
}
return output;
}
}
答案 0 :(得分:0)
首先将所有单词添加到字典中,然后再检查每个单词,因为您需要与文件中的所有单词进行比较 您采用的过程只是在比较之前先比较单词
import java.util.*;
import java.io.*;
public class CompositionTwo
{
private static List<String> dictionary = new ArrayList<>();
public static void main(String []args) {
File inputFile = new File("dictionary.txt");
File outputFile = new File("composedMore.txt");
Scanner in = null;
PrintWriter out = null;
String word;
try {
in = new Scanner(inputFile);
out = new PrintWriter(outputFile);
//Read the file indictionary
while (in.hasNext()) {
String input = in.next();
dictionary.add(input);
}
//Check All the words in dictionary for Splitting
for(int i=0;i<dictionary.size();i++)
{
String output = splitWord(dictionary.get(i), "");
if (!"".equals(output)) {
String outa[] = output.split("\\s") ;
if(outa.length >= 3) // Check if 3 words are created as output
{
System.out.println(dictionary.get(i) + ":" + output);
out.println(dictionary.get(i) + ":" + output);
}
}
}
} catch (IOException e) {
System.out.println(e.getMessage());
} finally {
in.close();
out.close();
}
}
public static String splitWord(String word, String output) {
if (word.length() == 0) {
return output;
}
else {
for (int i = 1; i <= word.length(); i++) {
// consider all prefixes of current String
String prefix = word.substring(0, i);
// if the prefix is present in the dictionary, add prefix to the
// output String and recurse for remaining String
if (dictionary.contains(prefix)) {
return splitWord(word.substring(i), output + " " + prefix);
}
}
}
return output ;
}
}
答案 1 :(得分:0)
假设您有这本字典:
disc
on
so
lateness
并假设您针对字符串disconsolateness
运行程序。
我进行了一些更改,首先加载字典,然后调用递归方法splitWord
。
您应该知道,递归方法被称为将全部放在堆栈中,然后逐个返回一个调用。因此,我选择将递归调用更改为:
return prefix + " " + splitWord(word.substring(i), output);
以便在每个方法调用之后将找到的每个单词连接起来。
很显然,就像您已经做过的那样,当变量output
中没有更多字符时,我的停止条件仍然存在。
package io.gdfb.questions;
import java.util.*;
import java.io.*;
public class CompositionTwo {
private static List<String> dictionary = new ArrayList<>();
public static void main(String[] args) {
loadDictionary();
File outputFile = new File("composedMore.txt");
PrintWriter out = null;
try {
out = new PrintWriter(outputFile);
String output = splitWord(args[0], "");
if (output != "") {
out.println(args[0] + ":" + output);
}
} catch (IOException e) {
System.out.println("An IOException has occurred during output process.");
} finally {
out.flush();
out.close();
}
}
private static void loadDictionary() {
InputStream inputFile = Thread.currentThread().getContextClassLoader().getResourceAsStream("dictionary.txt");
Scanner in = null;
try {
in = new Scanner(inputFile);
while (in.hasNext()) {
String input = in.next();
dictionary.add(input);
}
} finally {
in.close();
}
}
public static String splitWord(String word, String output) {
for (int i = 1; i <= word.length(); i++) {
// consider all prefixes of current String
String prefix = word.substring(0, i);
// if the prefix is present in the dictionary, add prefix to the
// output String and recurse for remaining String
if (dictionary.contains(prefix)) {
return prefix + " " + splitWord(word.substring(i), output);
}
}
return output;
}
}