Explanation on Integer.MAX_VALUE and Integer.MIN_VALUE to find min and max value in an array

but as for this method, I don't understand the purpose of Integer.MAX_VALUE and Integer.MIN_VALUE.

By starting out with smallest set to Integer.MAX_VALUE and largest set to Integer.MIN_VALUE, they don't have to worry later about the special case where smallest and largest don't have a value yet. If the data I'm looking through has a 10 as the first value, then numbers[i]<smallest will be true (because 10 is < Integer.MAX_VALUE) and we'll update smallest to be 10. Similarly, numbers[i]>largest will be true because 10 is > Integer.MIN_VALUE and we'll update largest. And so on.

Of course, when doing this, you must ensure that you have at least one value in the data you're looking at. Otherwise, you end up with apocryphal numbers in smallest and largest.

Note the point Onome Sotu makes in the comments:

...if the first item in the array is larger than the rest, then the largest item will always be Integer.MIN_VALUE because of the else-if statement.

Which is true; here's a simpler example demonstrating the problem (live copy):

public class Example{    public static void main(String[] args) throws Exception {        int[] values = {5, 1, 2};        int smallest = Integer.MAX_VALUE;        int largest  = Integer.MIN_VALUE;        for (int value : values) {            if (value < smallest) {                smallest = value;            } else if (value > largest) {                largest = value;            }        }        System.out.println(smallest + ", " + largest); // 1, 2 -- WRONG    }}

To fix it, either:

1. Don't use else, or

2. Start with smallest and largest equal to the first element, and then loop the remaining elements, keeping the else if.

Here's an example of that second one (live copy):

public class Example{    public static void main(String[] args) throws Exception {        int[] values = {5, 1, 2};        int smallest = values[0];        int largest  = values[0];        for (int n = 1; n < values.length; ++n) {            int value = values[n];            if (value < smallest) {                smallest = value;            } else if (value > largest) {                largest = value;            }        }        System.out.println(smallest + ", " + largest); // 1, 5    }}

Instead of initializing the variables with arbitrary values (for example int smallest = 9999, largest = 0) it is safer to initialize the variables with the largest and smallest values representable by that number type (that is int smallest = Integer.MAX_VALUE, largest = Integer.MIN_VALUE).

Since your integer array cannot contain a value larger than Integer.MAX_VALUE and smaller than Integer.MIN_VALUE your code works across all edge cases.

By initializing the min/max values to their extreme opposite, you avoid any edge cases of values in the input: Either one of min/max is in fact one of those values (in the case where the input consists of only one of those values), or the correct min/max will be found.

It should be noted that primitive types must have a value. If you used Objects (ie Integer), you could initialize value to null and handle that special case for the first comparison, but that creates extra (needless) code. However, by using these values, the loop code doesn't need to worry about the edge case of the first comparison.

Another alternative is to set both initial values to the first value of the input array (never a problem - see below) and iterate from the 2nd element onward, since this is the only correct state of min/max after one iteration. You could iterate from the 1st element too - it would make no difference, other than doing one extra (needless) iteration over the first element.

The only sane way of dealing with inout of size zero is simple: throw an IllegalArgumentException, because min/max is undefined in this case.