java - 如何将单行数字输入到数组中？

Question

我想像这样在一行中输入几个数字，

14 53 296 12 1

并用空格分隔数字，并将它们全部放在一个数组中。我该怎么做呢？另外，我如何确保他们输入的每个数字都是整数并且输入的数字少于 10 个？也许使用 try/catch 异常？

Answer 1

下面是一些代码来达到预期的效果：

int[] nums;
try {
    String line = ...; // read one line and place it in line
    StringTokenizer tok = new StringTokenizer(line);
    if (tok.countTokens() >= 10)
        throw new IllegalArgumentException(); // can be any exception type you want, replace both here and in the catch below
    nums = new int[tok.countTokens()];
    int i = 0;
    while (tok.hasMoreTokens()) {
        nums[i] = Integer.parseInt(tok.nextToken());
        i++;
    }
} catch (NumberFormatException e) {
    // user entered a non-number
} catch (IllegalArgumentException e) {
    // user entered more that 10 numbers
}

结果是nuns数组包含用户输入的所有整数。当用户键入非整数或超过 10 个数字时，将激活 catch 块。

Answer 2

读入行

String line = // how ever you are reading it in

按空间分割，查看文档String.split()

String[] numbers = line.split("\\s");

检查尺寸if(numbers.length > 10) //... to large

检查每个都是整数，看看Integer.parseInt()，然后放入你的新数组，所有这些都在一起......

 String line = //How you read your line
 String[] numbers = line.split("\\s");

 if(numbers.length <= 10)
 {
     int[] myNumbers = new int[numbers.length]
     int i = 0;
     for(String s:numbers) {
        try {
             int num = Integer.parseInt(s);
             myNumbers[i] = num;
             i++;
         } catch (NumberFormatException nfex) {
             // was not a number
         }
     }
  }
  else
      // To many numbers

Answer 3

String line = "12 53 296 1";
String[] line= s.split(" ");
int[] numbers = new int[splitted.length];
boolean correct=true;
if(splitted.length <10)
{
    correct=false;
}
for(int i=0;i<splitted.length;i++)
{
    try
    {
        numbers[i] = Integer.parseInt(splitted[i]);
    }
    catch(NumberFormatException exception)
    {
        correct=false;
        System.out.println(splitted[i] + " is not a valid number!");
    }
}

现在数组 numbers 包含解析的数字，并且布尔正确显示每个部分是否都是数字并且不少于 10 个数字。