Question

我试图将数字输入为string，然后将其拆分并存储在string数组中，然后再将它们作为整数存储在int数组中。我已经尝试了许多类似.trim()或scanner.skip()的方法，但是无法在此处解决此问题。

我的输入：

4 
1 2 2 2

 public static void main(String []args) throws IOException{

    Scanner scanner = new Scanner(System.in);
    int arCount = scanner.nextInt();
    int[] ar = new int[arCount];
    String[] arItems = scanner.nextLine().split(" ");

    for (int i = 0; i < arCount; i++) {
        int arItem = Integer.parseInt(arItems[i].trim());
        ar[i] = arItem;
    }
    scanner.close();
 }

收到的错误是：

Exception in thread "main" java.lang.NumberFormatException: For input string: ""
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:592)
at java.lang.Integer.parseInt(Integer.java:615)
at Hello.main(Hello.java:32)

Answer 1

您可以捕获异常并跳过不良记录。

我还建议您将一个或多个空格分开，以防止空字符串开头

Scanner scanner = new Scanner(System.in);
int arCount = scanner.nextInt();
scanner.nextLine();
int[] ar = new int[arCount];
String[] arItems = scanner.nextLine().split("\\s+");

for (int i = 0; i < arCount; i++) {
    try {
        ar[i] = Integer.parseInt(arItems[i].trim());
    } catch (NumberFormatException e) {
        System.err.println(arItems[i] + " is not an int");
    }
}

Answer 2

为什么不一次又一次地使用nextInt（）方法（完全跳过split的使用），就像这样，

import java.util.*;

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

    Scanner scanner = new Scanner(System.in);
    int arCount = scanner.nextInt();

    int[] ar = new int[arCount];
    for(int i=0;i<arCount;i++){
        ar[i] = scanner.nextInt();;
    }
    System.out.println(Arrays.toString(ar));

    scanner.close();
 }
}

更新：

在您的原始程序中，您首先使用nextInt（）方法，该方法不使用换行符'\ n'。因此，nextLine（）的下一次调用实际上消耗了同一行上的换行符，而不是包含数组整数的下一行。因此，为使代码正常工作，您可以按如下所示对其进行一些修改，

import java.util.*;

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

    Scanner scanner = new Scanner(System.in);
    int arCount = scanner.nextInt(); // doesn't consume \n
    int[] ar = new int[arCount];
    scanner.nextLine(); //consumes \n on same line 
    String line = scanner.nextLine(); // line is the string containing array numbers
    System.out.println(line);
    String[] arItems = line.trim().split(" "); //trim the line first, just in case there are extra spaces somewhere in the input
    System.out.println(Arrays.toString(arItems));
    for (int i = 0; i < arCount; i++) {
        int arItem = Integer.parseInt(arItems[i].trim());
        ar[i] = arItem;
    }
    System.out.println(Arrays.toString(arItems));
 }
}

Answer 3

也许可以在Scanner.nextLine（）之后尝试使用.trim（）。

function shuffleArray(array) {
    for (let i = array.length - 1; i > 0; i--) {
        const j = Math.floor(Math.random() * (i + 1));
        [array[i], array[j]] = [array[j], array[i]];
    }
}

function generateRandomNumbers(maxNumber, randomNumbersCount) {
  var possibleNumbers = [];
  // populate array with all possible values
  for (var i = 0; i <= maxNumber; i++) { possibleNumbers.push(i); }
  // shuffle the array to get a random order of the possible numbers O(n)
  shuffleArray(possibleNumbers);
  // trim the array down to only the first n numbers where n = randomNumbersCount
  possibleNumbers.length = randomNumbersCount;
  return possibleNumbers;
}

console.log (generateRandomNumbers(10, 5));
console.log (generateRandomNumbers(10, 5));
console.log (generateRandomNumbers(10, 5));

错误：线程“主”中的异常java.lang.NumberFormatException：对于输入字符串：“”

3 个答案: