我正在创建一个基于控制台的Evil Hangman程序。不是在游戏开头选择一个单词,而是通过选择包含用户输入的字母的单词来逐步缩小单词列表。
这是一项非常常见的任务,因此我确信您之前已经听说过类似的内容。
如果您需要更多说明,请参阅以下规范。
用于编译的必要文件和客户端:
此特定类中有问题的方法是record()方法,这是迄今为止在此分配中编程最困难的方法。 我观察到两个主要问题:
a)有以下NullPointerException,我无法找出原因。它发生在将一个单词添加到列表的行中。据我所知,单词和列表都应该存在并且都被初始化。这不是列表中没有包含单词或列表未正确初始化的问题。它似乎只发生在第二个用户输入之后。
Exception in thread "main" java.lang.NullPointerException
at HangmanManager.record(HangmanManager.java:66)
at HangmanMain.playGame(HangmanMain.java:59)
at HangmanMain.main(HangmanMain.java:39)
b)NullPointerException可能是因为我之前尝试修复另一个问题。在NullPointerException发生之前,我遇到了问题,确定程序选择的单词不符合预期输出的原因。无论我做了什么,程序在遍历结束时选择的单词总是" aa。"即使我故意选择了一个'作为猜测,应该有效地过滤掉了#a;"作为一种选择,它仍然出现了。
到目前为止,这是我的计划。
import java.util.*;
public class HangmanManager {
private String pattern;
private int length;
private int max;
private SortedSet<Character> guessesMade;
private Set<String> words;
private Set<String> currentWords;
private Map<String, Set<String>> patternMap;
public HangmanManager(List<String> dictionary, int length, int max){
if (length < 1 || max < 0){
throw new IllegalArgumentException();
}
this.length = length;
this.max = max;
words = new TreeSet<String>();
for (String word : dictionary){
if (word.length() == length){
words.add(word);
}
}
currentWords = new TreeSet<String>();
guessesMade = new TreeSet<Character>();
patternMap = new TreeMap<String, Set<String>>();
pattern = "";
for (int i = 0; i < length; i++){
pattern += "- ";
}
}
public Set<String> words(){
return words;
}
public int guessesLeft(){
return max - guessesMade.size();
}
public SortedSet<Character> guesses(){
return guessesMade;
}
public String pattern(){
if (words.isEmpty()){
throw new IllegalArgumentException("There are no words.");
}
return pattern;
}
public int record(char guess){
if (guessesLeft() < 1 || words.isEmpty()){
throw new IllegalStateException();
}
if (!words.isEmpty() && guessesMade.contains(guess)) {
throw new IllegalArgumentException();
}
guessesMade.add(guess);
int largestOccurences = 0;
for (String word: words){
System.out.println(word);
System.out.println(words.size());
if (patternMap.containsKey(pattern)){
System.out.println("patternMap contains pattern");
largestOccurences = generatePattern(word, guess);
currentWords.add(word);
currentWords = patternMap.get(pattern);
patternMap.put(pattern, currentWords);
} else {
currentWords.add(word);
patternMap.put(pattern, currentWords);
}
}
words = findFamily();
return largestOccurences;
}
private Set<String> findFamily(){
int maxSize = 0;
Map <String, Integer> patternCount = new TreeMap<String, Integer>();
for (String key : patternMap.keySet()){
patternCount.put(key, patternMap.get(key).size());
if (patternMap.get(key).size() > maxSize){
maxSize = patternMap.get(key).size();
pattern = key;
} else if (patternMap.get(key).size() == maxSize){
if (key.length() >= pattern.length()){
pattern = key;
maxSize = patternMap.get(key).size();
}
}
}
System.out.println("Current pattern: " + pattern);
return patternMap.get(pattern);
}
private int generatePattern(String s, char guess) {
int count = 0;
pattern = "";
for (int i = 0; i < length; i++){
if (s.charAt(i) == guess){
pattern += guess + " ";
count++;
} else {
pattern += "- ";
}
}
return count;
}
}
答案 0 :(得分:0)
您的问题出在patternMap
。
if (patternMap.containsKey(pattern))
{
System.out.println("patternMap contains pattern");
largestOccurences = generatePattern(word, guess);
//=> generatePattern changes the global variable pattern, so it may
//not be contained in patternMap anymore
currentWords.add(word);
currentWords = patternMap.get(pattern);//since pattern was changed,
//patternMap return null and
//you get NullPointerException on
//the next iteration of the loop
patternMap.put(pattern, currentWords);
}
当您致电pattern
时,您需要确保patternMap
中存在新更改的currentWords = patternMap.get(pattern);
。否则,您将获得null
。
答案 1 :(得分:0)
在分配值之前检查null。 currentWords = patternMap.get(pattern);
替换它 如果(patternMap.get(模式)!= NULL) currentWords = patternMap.get(pattern);
chk this out ...如果你有任何这样的任务,那么空检查......
祝你好运!