Question

我正在尝试在Java中实现一种简单的挂起情况，其中初始化一个字符数组，接受用户输入，并返回字符串输入中字母出现的所有索引。到目前为止，我有一个非常奇怪的输出，我认为这是由于我的初学者在我的循环和某些位置方面的失礼。如果您对这个问题有洞察力，请告诉我：

public class ArrayRandomString {
    public static void main(String[] args) {

        // Create an array of characters:
        Character[] anArray = { 'P', 'A', 'P', 'A', 'B', 'E', 'A', 'R' };
        for (char ch : anArray)
            System.out.print(ch + " ");
        System.out.println();

        System.out.println("This program initializes a secret word or phrase to be guessed by you!");
        System.out.println("Enter a string of less than 5 characters to return all indexes where it occurs in scrambled string: ");
        Scanner input = new Scanner(System.in);
        String string = input.next();

        System.out.println(string);
        System.out.println(Character.toUpperCase(string.charAt(0)));

        List<Integer> myList = new ArrayList<Integer>();
        List<Integer> noList = new ArrayList<Integer>();
        int i;
        int j;
        // Loop to find index of inputted character
        for (i = 0; i < anArray.length; i++) {
            Character ch = new Character(anArray[i]);
            for (j = 0; j < string.length(); j++) {
                List<Integer> letterList = new ArrayList<Integer>();
                if (Character.toUpperCase(string.charAt(j)) == ch) {
                    letterList.add(j);
                }
                System.out.println(Character.toUpperCase(string.charAt(j)) + " occurs at the following index " + letterList);
            }
        }
        // System.out.println("Your letter occurs at the following index in the scrambled array. No occurence, if empty: " + myList);
    }
}

如果用户输入为'PEAR'，理想输出：

P occurs at [0, 2]
E occurs at [5]
A occurs at [1, 3, 6]
R occurs at [7]

如果用户输入为'TEA'，理想输出：

T does not occur in string
E occurs at [5]
A occurs at [1, 3, 6]

到目前为止，我没有编写“不会发生”，因为“确实发生”已经很奇怪了。提前致谢

Answer 1

你的内部外部循环混淆了，你有时使用错误的数组/字符串使用错误的索引（i或j）。它有助于为i和j提供更具描述性的名称，以便您知道何时使用哪个。最后，你只能打印＆＃34;发生＆＃34;在您尝试查找所有匹配项之后的字符串，否则您将为每次出现打印整行，而不是为所有出现的字符打印一行。

这会修复你的循环：

for (int stringIndex = 0; stringIndex < string.length(); stringIndex++) {
    char ch = Character.toUpperCase(string.charAt(stringIndex));
    List<Integer> letterList = new ArrayList<Integer>();
    for (int anArrayIndex = 0; anArrayIndex < anArray.length; anArrayIndex++) {
        if (anArray[anArrayIndex] == ch) {
            letterList.add(anArrayIndex);
        }
    }
    System.out.println(ch + " occurs at the following index " + letterList);
}

我将此作为练习让自己测试letterList是否为空（提示：查看方法isEmpty()的Javadoc）并打印另一条消息以表明您没有＆＃ 39;找不到任何事件。

Answer 2

int i;
// Loop to find index of inputted character
// List to store the output
List<String> outList = new ArrayList<String> ();
char[] stringArray = string.toCharArray();    
for (char stringArrayChar: stringArray) {
    // StringBuilder to store output for each iteration
    StringBuilder sb = new StringBuilder();
    sb.append(Character.toUpperCase(stringArrayChar) + " occurs at ");

    // positions List
    List<Integer> posList = new ArrayList<Integer>();
    i = 0;
    for (char anArrayChar: anArray) {          
        if (anArrayChar == Character.toUpperCase(stringArrayChar)) {
           posList.add(i);              
        }
        i++;
    }
    // Check if the char is present in the specified array
    if (posList.size() > 0) { 
        sb.append(posList.toString());            
    } else {
        sb.append(" No Place in Array");
    }
    outList.add(sb.toString());
}
// Print the final result
for(String outString: outList ) {
    System.out.println(outString);
}

从用户输入字符串返回数组中的索引

2 个答案: