Question

我有一个简单的程序，要求您通过键盘输入数字（每个数字用空格或逗号分隔），然后将它们从低到高排序并打印出来。

问题是这些数字没有打印出来。

以下是源代码：

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

    Scanner scanner = new Scanner(System.in);
    System.out.print("Type random numbers: ");
    String input = new String(scanner.nextLine());
    scanner.close();
    String[] numString = new String[input.length()];

    int a = 0;
    int i = 0;
    for (; i < input.length() - 1;) {
        if (Character.isDigit(input.charAt(a))) { // If the character at input[a] is a digit
            numString[i] += Character.toString(input.charAt(a)); // it is added to numString[i]
            if(!(a+1 > input.length())){
                a++;
            }
        }
        if (numString[i] != null && !Character.isDigit(input.charAt(a))) { // If numString[i] is already in use and the char at input[a] is not a digit
            if(!(i+1 > input.length())){
                i++;
            }
            if(!(a+1 > input.length())){
                a++;
            }
        }
        if (numString[i] == null && !Character.isDigit(input.charAt(a))){ // If numString[i] is not in used and the character at char[a] is not a digit.
             if(!(a+1 > input.length())){
                a++;
             }
        }
    }
    a = 0;
    i = 0;
    int[] numbers = new int[numString.length];
    for(; i < numString.length - 1; i++){
        numbers[i] = Integer.parseInt(numString[i]);
    }
    quicksort(numbers, 0, numbers.length - 1);
    for(i = 0; i < numbers.length - 1; i++){
        if(i != numbers.length){
            System.out.print(numbers[i] + ", ");
        }else{
            System.out.println(numbers[i] + ".");
        }
    }
}

public static void quicksort(int numbers[], int left, int right) {

      int pivot = numbers[left]; // takes the first element as pivot
      int l = left; // l searches from left to right
      int r = right; // r searches from right to left
      int aux;

      while(l<r){ // While searches doesn't cross
         while((numbers[l] <= pivot) && (l < r)) l++; // searches for an element higher than the pivot
         while(numbers[r] > pivot) r--;         // Searches for an element smaller than the pivot
         if (l<r) {                      // if searches haven't been crossed                   
             aux = numbers[l];           // they are exchanged
             numbers[l] = numbers[r];
             numbers[r] = aux;
         }
       }
       numbers[left] = numbers[r]; // The pivot is placed in a way that we have the 
       numbers[r] = pivot;         // smaller digits at the left and the higher digits at the right
       if(left < r-1)
          quicksort(numbers, left, r-1); // left subarray is sorted
       if(r+1 < right)
          quicksort(numbers, r+1, left); // right subarray is sorted
    }
}

已编辑：在第26行添加了a++;表达式，阻止程序进入无限循环和新的if阻止程序阻止＃{1}} 34;冻结＆＃34;，但现在我收到了这个错误：

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 16
at java.lang.String.charAt(Unknown Source)
at StartHere.main(StartHere.java:21)

这是输入：

Type random numbers: 246, 421, 123, 2

Answer 1

我正在通过提供有效的替代方案来回答：

input.split("\\D+")

它只有一行。

根据大众的需求，以下是它的工作原理：

String[]返回\D个数字，由非数字分隔 - \D+表示“非数字”，Arrays.stream()表示一个或更多非数字
map(Integer::parseInt)从数组
String将每个Integer从拆分转为10 > 2。这是必需的，因此下一步将按数字顺序排序，而不是 lexographical 顺序 - n.b. "10" < "2"但sorted()
map(String::valueOf)对整数流进行排序（数字顺序）
Integers将Strings转回collect()，为下一步做好准备
Collectors.joining(", ", "", ".")将流合并到单个对象。请参阅下一点了解......
{{1}}是一个预先设定的收集器，它从流中生成一个String，并以参数1作为分隔符，参数2的前缀和参数3的后缀

Answer 2

当你的程序读取第一个＆＃34;，＆＃34; ，

它会增加＆＃34; i＆＃34;，但是＆＃34; a＆＃34;没有增加，

input.charAt（a）将是＆＃39;，＆＃39;和numString [i]为空

因此将跳过if条件并导致无限循环。

Answer 3

//Your Input String Is Like Type random numbers: 12,654,123,3 
//you first need to split this string with (,) because 
//scanner object read the line as string object so you need to fist split this input string with (,). then you can perform sort operation in easy way.

String[] splitStr = input.split(",");

Answer 4

以下是代码示例，该数字将打印出来。我修改了您的代码modified here。我希望它会对你有所帮助。

<强>编辑：

  public static void main(String[] args) {

    Scanner scanner = new Scanner(System.in);
    System.out.print("Type random numbers: ");
    String input = new String(scanner.nextLine());
    scanner.close();

    // modified here -> split whitespace, a literal comma
    String[] numString = input.split("\\s*,\\s*");

    int i = 0;
    int[] numbers = new int[numString.length];
    for (; i < numString.length; i++) {
        // modified here -> make sure numString[] isn't null and empty string
        if (numString[i] != null && !"".equals(numString[i])) {
            numbers[i] = Integer.parseInt(numString[i]);
        }
    }
    quicksort(numbers, 0, numbers.length - 1);
    // modified here
    for (i = 0; i < numbers.length; i++) {

        if (i < (numbers.length - 1)) {
            System.out.print(numbers[i] + ", ");
        } else {
            System.out.println(numbers[i] + ".");
        }
    }
}

Answer 5

试试这段代码。它只是做这项工作。您可以从逗号或空格分隔输入。但不是他们俩。

    Scanner scanner = new Scanner(System.in);
    System.out.print("Type random numbers: ");
    String input = new String(scanner.nextLine());
    scanner.close();
    String arr[];

    boolean isComma = false;

    //check for comma
    if (input.indexOf(",") != -1) {
        isComma = true;
    }

    //split based on comma or space
    if (isComma) {
        arr = input.split(",");
    } else {
        arr = input.split(" ");
    }

    int series[] = new int[arr.length];
    for (int i = 0; i < arr.length; i++) {
        series[i] = Integer.parseInt(arr[i].trim());
    }

    Arrays.sort(series);

    for (int i = 0; i < series.length; i++) {
        System.out.print(series[i] + " ");
    }

    System.out.println("");

我的程序没有打印出转换为整数的字符串

5 个答案: