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# Vector problem

Posted on 2004-04-17
Medium Priority
289 Views
Last Modified: 2012-05-04
I need to return the minimum and maximum INDEX of a vector:

public static int minimum(Vector costs)
{
int min = ((Integer) costs.get ( 0 ) ).intValue () ;
for ( int i = 1, iSize = costs.size () ; i < iSize ; i ++ )
{
Integer temp = ( Integer ) costs.get ( i ) ;
int tempVal = temp.intValue () ;

if ( tempVal < min )
min = i ;

}
return min;

}

public static int maximum(Vector costs)
{
int max = ((Integer) costs.get ( 0 ) ).intValue () ;
for ( int i = 1, iSize = costs.size () ; i < iSize ; i ++ )
{
Integer temp = ( Integer ) costs.get ( i ) ;
int tempVal = temp.intValue () ;

if ( tempVal > max )
max = i ;

}
return max;

}

the minimum works but not the maximum~
0
Question by:jtcy
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14 Comments

LVL 86

Expert Comment

ID: 10848993
Try

public static int maximum(Vector costs) {
int max = Integer.MIN_VALUE;
for (int i = 0, i < costs.size(); i++) {
Integer temp = (Integer) costs.get(i);
int tempVal = temp.intValue();
max = Math.max(max, tempValue);
}
return max;
}

0

LVL 86

Expert Comment

ID: 10848997
>>max = Math.max(max, tempValue);

should have been

max = Math.max(max, tempVal);
0

Author Comment

ID: 10849097
Thanks but I need to return the maximum INDEX, not value.
0

LVL 4

Expert Comment

ID: 10849100
Here's a general solution that will work with most any Vector where the class of all of the elements are the same, and implement comparable (true for your Integers, as well as Strings) --

import java.util.Vector;

public class VectorMinMax {

int minIndex = -1;
int maxIndex = -1;

public VectorMinMax(Vector v) {

if (v.size() == 0)
return;

Object minObject = v.get(0);
Object maxObject = v.get(0);

for (int i = 0; i < v.size(); i++) {
Comparable c = (Comparable) v.get(i);
if (c.compareTo(minObject) < 0) {
minObject = c;
minIndex = i;
}
if (c.compareTo(maxObject) > 0) {
maxObject = c;
maxIndex = i;
}
}
}

public int getMinIndex() {
return minIndex;
}

public int getMaxIndex() {
return maxIndex;
}

public static void main(String[] args) {
Vector v = new Vector();
v.add("One");
v.add("Two");
v.add("Alpha");
v.add("Beta");
VectorMinMax vmm = new VectorMinMax(v);
System.out.println("Min: " + v.get(vmm.getMinIndex()) + " at " + vmm.getMinIndex());
System.out.println("Max: " + v.get(vmm.getMaxIndex()) + " at " + vmm.getMaxIndex());

System.out.println();

v = new Vector();
v.add(new Integer(100));
v.add(new Integer(-10));
v.add(new Integer(1));
v.add(new Integer(400));
v.add(new Integer(10000));
vmm = new VectorMinMax(v);
System.out.println("Min: " + v.get(vmm.getMinIndex()) + " at " + vmm.getMinIndex());
System.out.println("Max: " + v.get(vmm.getMaxIndex()) + " at " + vmm.getMaxIndex());
}
}

And here's the run:

Min: Alpha at 2
Max: Two at 1

Min: -10 at 1
Max: 10000 at 4
0

LVL 4

Expert Comment

ID: 10849107
So bascially you would take your vector costs and just go

VectorMinMax vmm = VectorMinMax(costs);
int min = vmm.getMinIndex();
int max = vmm.getMaxIndex();

Etc.

If you want to go whole hog, you could extend Vector and override get(), add(), etc., and keep the min/max as you add/remove elements from your Vector, which would be pretty neat, and a lot more efficient.
0

LVL 86

Expert Comment

ID: 10849108
>>Thanks but I need to return the maximum INDEX, not value.

I don't know what you mean by that. If you mean as follows

index 0 - value 3948593485
index 1 - value 3948579
index 2 - value 928759

then the maximum index is given by

int maxIndex = v.size() - 1;
0

LVL 4

Expert Comment

ID: 10849111
He means:

What is the index of the max value?
What is the index of the min value?
0

LVL 86

Expert Comment

ID: 10849115
Oh I see! You mean the index of the maximum value. That simply is the following - using my original code

int indexOfMaximumValue = v.indexOf(new Integer(maximum(v)));
0

LVL 86

Expert Comment

ID: 10849119
And of course if you sort the Vector:

int indexOfMinimumValue =  0;
int indexOfMaximumValue = v.size() - 1;
0

LVL 4

Expert Comment

ID: 10849122
jtcy -- one thing you should decide is what the "max index" is when more than one Vector element have the same value.
0

LVL 86

Expert Comment

ID: 10849125
So the simplest thing to do if you don't want to disturb the sort order is to do:

Vector v2 = (Vector)v.clone();
Collections.sort(v2);
int indexOfMinimumValue =  0;
int indexOfMaximumValue = v2.size() - 1;
0

LVL 4

Expert Comment

ID: 10849137
CEHJ: You probably mean: int indexOfMinimumValue = v.indexOf(v2.get(0)); // etc.
0

LVL 86

Expert Comment

ID: 10849147
Yes.

Since you haven't explained yourself very well jtcy, i completely misunderstood your requirement. You should use john's answer
0

LVL 92

Accepted Solution

objects earned 1000 total points
ID: 10851169
public static int minimum(Vector costs)
{
int min = ((Integer) costs.get ( 0 ) ).intValue () ;
int mini = 0;
for ( int i = 1, iSize = costs.size () ; i < iSize ; i ++ )
{
Integer temp = ( Integer ) costs.get ( i ) ;
int tempVal = temp.intValue () ;

if ( tempVal < min )
{
mini = i ;
min = tempVal;
}
}
return mini;

}

public static int maximum(Vector costs)
{
int maxi = 0;
int max = ((Integer) costs.get ( 0 ) ).intValue () ;
for ( int i = 1, iSize = costs.size () ; i < iSize ; i ++ )
{
Integer temp = ( Integer ) costs.get ( i ) ;
int tempVal = temp.intValue () ;

if ( tempVal > max )
{
maxi = i ;
max = tempVal;
}

}
return maxi;

}
0

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