Fastest way of finding the middle value of a triple?

Question

Given is an array of three numeric values and I'd like to know the middle value of the three. The question is, what is the <i>fastest</i> way of finding the middle of the three? My approach is this kind of pattern - as there are three numbers there are six permutations: if (array[randomIndexA] >= array[randomIndexB] && array[randomIndexB] >= array[randomIndexC]) It would be really nice, if someone could help me out finding a <i>more elegant</i> and <i>faster</i> way of doing this.

Answer 1

I would just bubble sort the three....

#include <stdio.h>

Swap( int * X, int * Y)

{

int temp = *Y;

*Y = *X;

*X=temp;

}

BubbleSort3( int* A)

{

if (A[0]>A[1]) { Swap(&A[0],&A[1]); }

if (A[1]>A[2]) { Swap(&A[1],&A[2]); }

if (A[0]>A[1]) { Swap(&A[0],&A[1]); }

}

int main()

{

int A[3];

//change these to whatever you like; Or you can have the use INPUT the values, read them from file, input from command line, etc.

A[0]=13;

A[1]=7;

A[2]=10;

BubbleSort3(A);

printf(" the median is %d \n",A[1]);

}

===================================================

Sorry you asked for Java.... but still the same reasoning and logic...

bubble sort the three integers and return A[1] as the median...

Here's some quick and dirty java

class MedianOf3

{

public int getMedian(int [] A)

{

int temp;

if (A[0]>A[1])

{

temp = A[1];

A[1] = A[0];

A[0] = temp;

}

if (A[1]>A[2])

{

temp = A[2];

A[2] = A[1];

A[1] = temp;

}

if (A[0]>A[1])

{

temp = A[1];

A[1] = A[0];

A[0] = temp;

}

return(A[1]);

}

public static void main(String args[])

{

MedianOf3 myMedian = new MedianOf3();

int A[] = {7,13,10};

System.out.println(" Median is " + myMedian.getMedian(A));

}

}