Java reverse an int value without using array

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Java reverse an int value – Principles

  1. Modding (%) the input int by 10 will extract off the rightmost digit. example: (1234 % 10) = 4

  2. Multiplying an integer by 10 will “push it left” exposing a zero to the right of that number, example: (5 * 10) = 50

  3. Dividing an integer by 10 will remove the rightmost digit. (75 / 10) = 7

Java reverse an int value – Pseudocode:

a. Extract off the rightmost digit of your input number. (1234 % 10) = 4

b. Take that digit (4) and add it into a new reversedNum.

c. Multiply reversedNum by 10 (4 * 10) = 40, this exposes a zero to the right of your (4).

d. Divide the input by 10, (removing the rightmost digit). (1234 / 10) = 123

e. Repeat at step a with 123

Java reverse an int value – Working code

public int reverseInt(int input) {
    long reversedNum = 0;
    long input_long = input;

    while (input_long != 0) {
        reversedNum = reversedNum * 10 + input_long % 10;
        input_long = input_long / 10;
    }

    if (reversedNum > Integer.MAX_VALUE || reversedNum < Integer.MIN_VALUE) {
        throw new IllegalArgumentException();
    }
    return (int) reversedNum;
}

You will never do anything like this in the real work-world. However, the process by which you use to solve it without help is what separates people who can solve problems from the ones who want to, but can’t unless they are spoon fed by nice people on the blogoblags.

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