Question

我在comp 182课程中有一个项目，我们正在研究向量，但我仍然坚持制作一个“有序向量”。当我尝试运行它时，我得到一个ArrayOutofBounds错误。

（“howMany”变量是数组“theWords”中字符串大小的计数并且代码从另一个类中运行，该类读取其中包含10个单词的输入文件，使用此“addWord”方法将文件中的单词添加到“theWords”数组中。）

这是我到目前为止的代码：[顺便说一句，我们不允许只使用“数组”方法“compareTo”]

 public void addWord(String newWord) {
  //adds in words
     if (howMany < theWords.length) {
        theWords[howMany]= newWord; 
        howMany++;
     }  
     else {
        String t [] = new String[capacity+10];
        for (int i=0; i <capacity; i++){
           t[i] = theWords[i];
        }
        theWords = t;
        theWords[howMany] = newWord; 
        howMany++;
     }

    //ordering words
     for(int g = howMany - 1, z = howMany ; g < howMany; g--, z--) {
        if(newWord.compareTo(theWords[g]) < 0) {
           theWords[g] = theWords[z];
           theWords[g] = newWord;
        }
            else
            newWord = theWords[z];          
     }
        howMany++;
  }

非常感谢任何帮助！

Answer 1

取决于数组的容量，语句

theWords[g] = theWords[z]

将在第一次执行循环时失败，因为z =数组的长度（数组中的最后一个索引是length - 1）。在旁注中，在循环中，您最初设置

g = howMany - 1

然后递减它，所以g＆lt; howMany将永远是真的...你将有一个无限循环。当g或z低于0时，此无限循环也可能导致索引超出范围异常。

Answer 2

如果在执行函数之前howMany == capacity - 1，则会出现问题。

会发生什么：你进入if的第一个分支（//添加单词）然后你增加howMany，然后你成为howMany = capacity然后你分配z = howMany并访问一个外面的数组它的界限：theWords[g] = theWords[z];。

另一件事是你没有增加capacity变量，它必须是capacity+=10。

这是我的变体：

import java.util.Arrays;
import java.util.Random;
public class OrderedArray{

  int howMany = 0;
  String[] theWords = new String[10];

  public void addWord(String newWord) {

    //don't accept nulls
    if(newWord == null) {
      return;
    }
    //if length is reached increase the array
    if(howMany >= theWords.length - 1) {
      theWords = Arrays.copyOf(theWords, theWords.length+10);
    }

    //go through existing words and add newWord if it is less then met value
    boolean isAdded = false;
    for(int idx = 0; idx < howMany && theWords[idx] != null; idx++){                                                          
      if(newWord.compareTo(theWords[idx]) < 0){
        isAdded = true;
        String valToShift = theWords[idx];
        theWords[idx] = newWord; 

        //copy all values after the met index
        for(int shIdx = idx+1; shIdx <= howMany && shIdx < theWords.length; shIdx++){
          String tmp = theWords[shIdx];
          theWords[shIdx] = valToShift;
          valToShift = tmp;
        }        
        break;
      }
    }

    //if a value was not added then add it
    if(!isAdded){
      theWords[howMany] = newWord;
    }
   ++howMany;
  }

  public String toString(){
    StringBuffer sb = new StringBuffer("howMany:");
    sb.append(howMany);
    for(int idx = 0; idx < howMany; idx++){
      sb.append(",");
      sb.append(theWords[idx]);
    }
    return sb.toString();
  }


  public static void main(String[] args){
    OrderedArray m = new OrderedArray();
    Random r = new Random(System.currentTimeMillis());
    for(int i = 0; i < 210; i++){
      m.addWord(Math.abs(r.nextInt())+ "");
    }
    System.out.println(m);

    OrderedArray m1 = new OrderedArray();

    m1.addWord("c");
    m1.addWord("d");
    m1.addWord("a");
    m1.addWord("cd");

    System.out.println(m1);

  }

}

Java - 排序/字母顺序字符串数组

2 个答案: