Question

我一直在

Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 3, Size: 3
at java.util.ArrayList.rangeCheck(Unknown Source)
at java.util.ArrayList.get(Unknown Source)
at hartman.ShortestString.printShortestString(ShortestString.java:40)
at hartman.ShortestString.main(ShortestString.java:28)

我该如何解决这个问题？

package hartman;

import java.util.ArrayList;
import java.util.Scanner;

public class ShortestString {

    public static void main(String[] args) {
        System.out.printf("WELCOME TO SHORTEST STRING\n\n");
        System.out.printf("Type \".\" when done entering data.\n\n");

        ArrayList<String> myArray = new ArrayList<>();
        Scanner keyboard = new Scanner(System.in);
        boolean keepAsking = true;

        while (keepAsking) {
            System.out.printf("Enter string: ");
            String userInput = keyboard.nextLine();

            if (userInput.equals(".")) {
                keepAsking = false;
            } else {
                myArray.add(userInput);
            }
        }

        printShortestString(myArray);
        System.out.printf("\n\nGOODBYE!\n");
        keyboard.close();
    }

    public static void printShortestString(ArrayList<String> myArray) {

        int index;
        int index1 = 1;

        for (index = 0; index < myArray.get(index).length(); index++) {
            if (myArray.get(index).length() < myArray.get(index1).length()) {
                System.out.printf("\nShortest string is \"%s\" with length %d",
                        myArray.get(index), myArray.get(index).length());
            } else {
                index1++;
            }
        }
        return;
    }
}

Answer 1

尝试将for (index = 0; index < myArray.length(); index++) {用于第40行。您使用元素index处字符串的长度，而不是ArrayList的长度。

Answer 2

你的方法似乎有问题。

public static void printShortestString(ArrayList<String> myArray) {
    if (myArray.isEmpty()) return;
    int len = myArray.get(0).length();
    int shortestIndex = 0;
    for (int index = 1; index < myArray.size(); index++) {
        if (myArray.get(index).length() < myArray.get(index - 1).length()) {
            len = myArray.get(index).length();
            shortestIndex = index;
        } 

    }
    System.out.printf("\nShortest string is \"%s\" with length %d",
                myArray.get(shortestIndex), len);
    return;
}

Answer 3

我们需要查看printShortestString()的来源，但我敢打赌，当索引i小于1时，您需要更改for循环。数组的长度，而不是等于数组的长度：

for {i=0; i < myArray.length(); i++) {
...
}

Answer 4

此行导致您退出索引：

for (index = 0; index < myArray.get(index).length(); index++)

您希望迭代数组的大小：

for (index = 0; index < myArray.size(); index++)

Answer 5

printShortestString(...)背后的逻辑不正确。您希望遍历列表大小（而不是字符串大小）。这里发生的事情是：

myArray   .get(index)                             .length()
list      object at a given index of this list    size of that object

你想要的是：

myArray  .size()
list     size of list

变化：

for (index = 0; index < myArray.get(index).length(); index++) {

要：

for (index = 0; index < myArray.size(); index++) {

Answer 6

String word = null;
for (index = 0; index < myArray.size(); index++) {

    if (word == null) {
        word = myArray.get(index);
        continue;
    }

    if (myArray.get(index).length() < word.length()) {
        word = myArray.get(index).length()
    }
}
System.out.println("The shortest word is:" + word );

java.lang.IndexOutOfBoundsException：索引：3，大小：3

6 个答案: