如何在Java中找到正确的数据类型来存储给定的整数？

Question

假设给了我一个数字，并要求我找出哪种数据类型适合存储它。例如，假设这个数字是741，那么它可以存储在short、int以及long数据类型中。但是，如果输入的数字很大，那么它只能存储在长数据类型中。根据用户输入的数字，我需要提供可用于存储它的所有可能的数据类型。

PS:这是一个在带有Java数据类型的Hackerrank平台上出现的问题。问题链接：- https://www.hackerrank.com/challenges/java-datatypes/problem?h_r=next-challenge&h_v=zen&isFullScreen=false

Answer 1

您可以使用原始数据类型包装器，它们都包含一个常量值和它们可以保持的max和min值。

它们是：

Byte.MIN_VALUE; //-128

Byte.MAX_VALUE; //127

Short.MIN_VALUE; //-32768

Short.MAX_VALUE; //32767

Integer.MIN_VALUE; //0x80000000

Integer.MAX_VALUE; //0x7fffffff

Long.MIN_VALUE; //0x8000000000000000L

Long.MAX_VALUE; //0x7fffffffffffffffL

这是我的解决方案：

    long n = 0;
    try {
        n = new Scanner(System.in).nextLong();
    } catch (InputMismatchException e) { //we're expecting just numbers anyway
        System.out.println("doesn't fit anywhere");
    }
    
    if (n > 0) {
        //since n is positive,
        //first let's evaluate if it's not too big to fit in a byte data type
        if (n <= Byte.MAX_VALUE) System.out.println("fits in a byte");
        
        //now let's evaluate if n is not too big to fit in a short data type
        if (n <= Short.MAX_VALUE) System.out.println("fits in a short");
        
        //now let's evaluate if n is not too big to fit in an int data type
        if (n <= Integer.MAX_VALUE) System.out.println("fits in an int");
        
        
    } else {
        //similar code for negative numbers
        if (n >= Byte.MIN_VALUE) System.out.println("fits in a byte");
        if (n >= Short.MIN_VALUE) System.out.println("fits in a short");
        if (n >= Integer.MIN_VALUE) System.out.println("fits in an int");
    }
    //well, since we already verified n doesn´t overtake a long with try block
    // n will always fit in a long
    System.out.println("fits in a long");