如何在Java中比较INT值中的数字？

Question

I'm writing a code for an assignment for school but it requires me to compare two int values. For example, if you have the number 123 and the other number is 321, they have the same digits but are in different orders. Is there easy way of comparing them or do i have to make them into a string and compare them as string types? if its the latter, how could i compare two strings? Is there any way of doing this without an array?

Answer 1

By comparing int values, if you mean greater than, less than or equal you can do that like so.

int a = 123, b= 321;

if(a > b) 
//a is greater than b (b is less than a)

if(a == b)
// a is equal to b

if(a < b)
// a is less than b (b is greater)

Could use some clarification, if you want to check if the number is reversed like you said in an example its called a palindrome.

You could reverse a number in the following if you had experience with loops and modulo(the %) in the following snippet.

int r = 0;
while(number != 0){
 r = r * 10 + number % 10;
 number /= 10;  }
 return r;

r would be that number reversed. If you input let's say 123 you would get 321 back, then you could compare it to the other to see if its just the reverse.

Let me know if you have any more questions and I'll try to answer!

To check if a number is arbitrarily mixed and not reversed to winning number, you could try the following.

Two numbers a and b, a is the winning number and b is the number the user chose. a is 251 and b is 521.

You could do this on each number to separate them.

int p1,p2,p3;
p1 = num % 10;
p2 = num / 10 % 10;
p3 = num / 100 % 10;

This would separate ex. 251 into 2, 5, and then 1. Then you could add them as so doing the same process for the second. sum is p1 + p2 + p3 and sum2 is p4 + p5 + p6 for the second number. Provided the numbers are not reversed. Use the thing I mentioned before for that case to check if they are flipped.

if(sum == sum2)
//Numbers are mixed but you won!

This should work.

Answer 2

Certainly not the fastest solution, but the code is short and easy to understand.

public boolean arePalindromes(int a, int b){
    //convert them to char arrays for easy sorting
    char[] aryA = String.valueOf(a).toCharArray();
    char[] aryB = String.valueOf(b).toCharArray();

    //sort them
    Collections.sort(aryA);
    Collections.sort(aryB);

    //put them back to strings for easy comparison
    String strA = new String(aryA);
    String strB = new String(aryB);

    //compare
    return strA.equals(strB);
}

Answer 3

Please try the following code (I have tested it), of which idea is borrowed from here and here . This solution still uses array.

public class CompareInt {
    public static void main(String[] args) {
        System.out.println(containSameDigits(123, 123));
        System.out.println(containSameDigits(123, 321));
        System.out.println(containSameDigits(123, 132));
        System.out.println(containSameDigits(123, 323));
        System.out.println(containSameDigits(123, 124));
        System.out.println(containSameDigits(123, 111));
    }

    public static boolean containSameDigits(int x, int y) {
        String xSorted = getSortedString(x);
        String ySorted = getSortedString(y);
        return xSorted.equalsIgnoreCase(ySorted);
    }

    public static String getSortedString(int x) {
        String xSorted = "";
        for (int digit = 0; digit < 9; digit++) {
            for (int temp = x; temp > 0; temp /= 10) {
                if (temp % 10 == digit) {
                    xSorted += digit;
                }
            }
        }
        return xSorted;
    }
}

Output:

true
true
true
false
false
false