我的程序不打印轉換為整數的字符串

Question

我有一個簡單的程序，要求您通過鍵盤輸入數字（每個數字用空格或逗號分隔），然后將它們從低到高排序並打印出來。

問題是數字沒有打印出來。

這是源代碼：

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
    }
}

編輯：添加了a++; 表達式在第26行阻止程序進入無限循環和新的if塊阻止程序“凍結”，但現在我收到此錯誤：

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)

這是輸入：

Type random numbers: 246, 421, 123, 2

Answer 1

我正在通過提供一種有效的替代方案來回答：

System.out.println( Arrays.stream(input.split("\\D+"))
  .map(Integer::parseInt)
  .sorted()
  .map(String::valueOf)
  .collect(Collectors.joining(", ", "", ".")));

它只有1行。

---

根據大眾需求，以下是它的工作原理：

• input.split("\\\\D+")返回數字的String[] ，用非數字分隔 - \\D表示“非數字”， \\D+表示一個或多個非數字
• Arrays.stream()從數組創建流
• map(Integer::parseInt)將每個String從拆分轉換為Integer 。 這是必要的，所以下一步將按數字順序排序，而不是排序順序 - nb 10 > 2但是"10" < "2"
• sorted()對整數流進行sorted() 數字順序排列）
• map(String::valueOf)將Integers轉回Strings ，為下一步做好准備
• collect()將流組合到單個對象。 請參閱下一點了解......
• Collectors.joining(", ", "", ".")是一個預先固定的收集器，它從流中生成一個String，將它們與參數1作為分隔符，參數2的前綴和參數3的后綴連接起來

Answer 2

當你的程序讀取第一個“，”時，

它將增加“i”，但“a”不會增加，所以，

input.charAt（a）將為'，'且numString [i]為null

因此，將跳過if條件並導致無限循環。

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

這是代碼示例，該數字將打印出來。 我修改了你modified here修改的代碼。 我希望它會對你有所幫助。

編輯：

  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

試試這個代碼。 它只是做這項工作。 您可以從逗號或空格分隔輸入。 但不是他們倆。

    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("");