如何使用条件运算符获取五个变量中的最高值和最低值？

Question

So far, I've only figured out how to get the highest and lowest values from the three variables using the conditional operator. Basically, this is what it looks like on the three variables:

int highnum = n1>n2?n1:n2; highnum = n3>highnum?n3:highnum;

Same concept goes to finding the lowest which is just replacing (>) with (<) . When it comes to five variables, I still can't quite know how to write them.

Answer 1

You can do so by putting the second condition in the else part (ie the part which after : ).

Demo :

public class Main {
    public static void main(String[] args) {
        int x = 15, y = 5, z = 10;
        int max = x > y && x > z ? x : y > x && y > z ? y : z;
        int min = x < y && x < z ? x : y < x && y < z ? y : z;

        System.out.println("Max: " + max + ", Min: " + min);
    }
}

Output :

Max: 15, Min: 5

If you are prohibited from using && , you can do it as follows:

int max = x > y ? (x > z ? x : z) : y > z ? y : z;
int min = x < y ? (x < z ? x : z) : y < z ? y : z;

Answer 2

Starting Point

Let's assume that the variables are

• n1
• n2
• n3
• n4
• n5

Doing it with ternary operator:

max = (max = (max = (max = (n1 > n2 ? n1 : n2)) > n3 ? max : n3) > n4 ? max : n4) > n5 ? max : n5;

This is incredibly messy (there might be a typo in the untested code above).

Doing it with conditionals

max = n1;
if (n2 > max) max = n2;
if (n3 > max) max = n3;
if (n4 > max) max = n4;
if (n5 > max) max = n5;

Doing it with arrays

int myArray = new int[] {n1, n2, n3, n4, n5};
int max = n1;
for (index = 1; index < myArray.length; index++) {
    if (max < myArray[index]) max = myArray[index];
}

Doing it with Math.max

int max = Math.max(Math.max(Math.max(Math.max(n1, n2), n3), n4), n5)

IntStream

As @Edwin Dalorzo pointed out in the comment section, IntStream is also a possible approach. This is the first time I have heard of IntStream , so the following example may be incorrect. If so, I graciously thank any issue pointed out about it:

IntStream.of(n1,n2,n3,n4,n5).reduce(max,Math::max)

Explanation (as far as I understand):

• we instantiate a stream by passing the 5 variables we have ( https://docs.oracle.com/javase/8/docs/api/java/util/stream/IntStream.html#of-int...- )
• we call reduce , so, under the hood all elements are considered and the operator is applied ( https://docs.oracle.com/javase/8/docs/api/java/util/stream/IntStream.html#reduce-int-java.util.function.IntBinaryOperator- )
• we pass max , a variable as first parameter to reduce , to specify that we compute the maximum
• we pass Math::max as the operator, so reduce will compute the maximum.

Answer 3

The optimal way is to use a loop/ define you own max function with Varargs¹ .

It's very nasty to do it using only conditional operators, if you need it for an exercise you can do this:

int highnum = ((n1>n2 ? n1 : n2) > (n3>n4 ? n3 : n4) ? (n1>n2 ? n1:n2) : (n3>n4?n3:n4)) > n5 ? ((n1>n2 ? n1 : n2) > (n3>n4 ? n3 : n4) ? (n1>n2 ? n1:n2) : (n3>n4?n3:n4)) : n5;

Or you can define a method as following:

public static int max(int a,int b){
    return a>b?a:b;
}

and then do something like this:

int highnum = max(max(max(n1,n2),max(n4,n3)),n5);

1: Varargs method:

public static int max(int... nums){
    int max = nums[0];
    for (int num : nums) if(num>max) max=num;
    return max;
}

int highnum = max(n1,n2,n3,n4,n5);

Answer 4

Anyway, don't feel it makes sense to do it with ternary operator for 5 variables.(Unless you want to train your brain).

Below just supplement on how to do it using Stream.

Stream approach

Make use of IntStream and corresponding summaryStatistics method:

import java.util.IntSummaryStatistics;
import java.util.stream.IntStream;

public class IntStatistic {
    public static void main(String[] args) {
        int n1 = 9;
        int n2 = 7;
        int n3 = 5;
        int n4 = 3;
        int n5 = 1;
        IntSummaryStatistics statistic = IntStream.of(n1, n2, n3, n4, n5).summaryStatistics();
        System.out.println("min " + statistic.getMin());
        System.out.println("max " + statistic.getMax());
    }
}