Find Even/Odd number without using mathematical/bitwise operator in Java

Question

如何在不使用数学/按位运算符的情况下找到偶数/奇数？

Answer 1

Are you allowed to cheat by going to a simpler, rather than trickier, solution than the standard i % 2 == 0 ? :) And secondarily, are you allowed to cheat by calling a JDK method which uses arithmetic or bitwise ops? If so, you could:

• get the int as a String
• look at the last char
• use that char in a switch
• return true iff that last char is '0', '2', '4', '6' or '8'.

Answer 2

How about this:

    int value = 7;

    String message = String.format("Is %s odd or even?",value);
    Object[] options = {"Odd","Even"};
    int n = JOptionPane.showOptionDialog(null,
            message,
            "Is It Odd Or Even",
            JOptionPane.YES_NO_OPTION,
            JOptionPane.QUESTION_MESSAGE,
            null,
            options,
            options[0]);
    System.out.println(String.format("%s is %s", value, options[n]));

The original question doesn't give the size of the data set.

Answer 3

This seems like a really weird arbitrary condition but what the heck. Just convert it to a string and check if the last character is a 0, 2, 4, 6, or 8.

Answer 4

I believe the answer is, you can't.

Any solution might consider will be composed of bitwise or mathematical operators. You can call or write a method which hides this from you, but it uses operators all the same.

---

you can do

public static boolean isOdd(int n) {
    return (n << -1) < 0;
}

Answer 5

I can think of the following ways to check if a number is odd, listed in order of performance:

// Using bit-manipulation (fastest)
boolean odd = ((num & 1) == 1);

// Using remainder (fast, but division is slower than bit-manipulation)
boolean odd = ((num % 2) == 1);

// Using conversion to string (very slow, but doesn't use any operators)
boolean odd = Integer.toString(num, 2).endsWith("1");

Answer 6

Try below java code.

String[] sarr = {"Odd Number","Even Number"};

int num = 7;

int x = num%2;

System.out.println(sarr[x]);