More convenient way to find the max of 2+ numbers?

Question

I want code that accepts more than 2 integers and prints out the biggest one. I used Math.MAX but the problem is that it accepts only 2 integers by default, and you can't print all the ints in it. So I had to make it like this:

int max = Math.max(a, Math.max(b, Math.max(c, Math.max(d, e))));

Is there a better method to do this?

Always include all relevant code **in the question itself**, don't just link. Why: meta.stackoverflow.com/questions/118392/add-stack-overfow-faq-entry-or-similar-for-putting-code-in-the-question — T.J. Crowder, Mar 30 '13 at 10:54
Try to use bitwise operator as you can see in http://stackoverflow.com/questions/4799287/maximum-of-two-number-in-c-using-bitwise-operators — Mihai8, Mar 30 '13 at 11:04
@user1929959: Why would you *want* to do that? Sounds like a far less readable solution to me... — Jon Skeet, Mar 30 '13 at 11:05
@Jon Skeet Using bitwise operators can do it faster than using Math function. — Mihai8, Mar 30 '13 at 11:09

NPE · Accepted Answer · 2013-03-30T11:14:23.573

9

You could use varargs:

public static Integer max(Integer... vals) {
    Integer ret = null;
    for (Integer val : vals) {
        if (ret == null || (val != null && val > ret)) {
            ret = val;
        }
    }
    return ret;
}

public static void main(String args[]) {
    System.out.println(max(1, 2, 3, 4, 0, -1));
}

Alternatively:

public static int max(int first, int... rest) {
    int ret = first;
    for (int val : rest) {
        ret = Math.max(ret, val);
    }
    return ret;
}

edited Mar 30 '13 at 11:14

answered Mar 30 '13 at 10:56

NPE

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I really can't understand what this does, can you please explain? – Sartheris Stormhammer Mar 30 '13 at 10:57
12

Maybe look at it for more than 5 seconds and try to understand. – Boris the Spider Mar 30 '13 at 10:59
Any reason for using `Integer` here rather than `int`? – Jon Skeet Mar 30 '13 at 11:04
@JonSkeet: It *returns* `Integer` mainly for the convenience of being able to return `null` when the input array is empty. And it *takes* `Integer` for consistency with the return type. – NPE Mar 30 '13 at 11:06
@NPE: Given that the OP only needs it to take "more than 2", why not make a non-varargs first `int` parameter, then a varargs remaining parameter, and make everything `int`. – Jon Skeet Mar 30 '13 at 11:10
@JonSkeet: That too would work, and would simplify the code. – NPE Mar 30 '13 at 11:13
@JonSkeet: I've added an `int`-based implementation to my answer. Thanks for the suggestion. – NPE Mar 30 '13 at 11:14

score 1 · Answer 2 · answered Mar 30 '13 at 10:57

1

You can use a simple loop:

public Integer max(final Collection<Integer> ints) {
    Integer max = Integer.MIN_VALUE;
    for (Integer integer : ints) {
        max = Math.max(max, integer);
    }
    return max;
}

answered Mar 30 '13 at 10:57

Boris the Spider

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As explained by http://stackoverflow.com/a/9303923/14731 you should use `Integer.NEGATIVE_INFINITY` instead of `Integer.MIN_VALUE`: `Double.MIN_VALUE` returns "the smallest positive nonzero value of type double", which is a number very close to 0, NOT the largest possible negative double. Instead, something like `Double.NEGATIVE_INFINITY` or `-Double.MAX_VALUE` could be used. – Gili Nov 23 '15 at 00:10
1

@Gili No. `Double.MIN_VALUE` is **very** different to `Integer.MIN_VALUE`. `Integer.NEGATIVE_INFINITY` does not exist. Your comment is wrong - please read the documentation before posting. – Boris the Spider Nov 23 '15 at 00:13
Apologies, you are right. – Gili Nov 23 '15 at 16:47

More convenient way to find the max of 2+ numbers?

2 Answers2