### Quicksort implementation by using Java

source: http://www.algolist.net/Algorithms/Sorting/Quicksort.

The divide-and-conquer strategy is used in quicksort. Below the recursion step is described:
1st: Choose a pivot value. We take the value of the middle element as pivot value, but it can be any value(e.g. some people would like to pick the first element and do the exchange in the end)

2nd: Partition. Rearrange elements in such a way, that all elements which are lesser than the pivot go to the left part of the array and all elements greater than the pivot, go to the right part of the array. Values equal to the pivot can stay in any part of the array. Apply quicksort algorithm recursively to the left and the right parts - the previous pivot element excluded!

Partition algorithm in detail:

There are two indices i and j and at the very beginning of the partition algorithm i points to the first element in the array and j points to the last one. Then algorithm moves i forward, until an element with value greater or equal to the pivot is found. Index j is moved backward, until an element with value lesser or equal to the pivot is found. If i ≤ j then they are swapped and i steps to the next position (i + 1), j steps to the previous one (j - 1). Algorithm stops, when i becomes greater than j.

After partition, all values before i-th element are less or equal than the pivot and all values after j-th element are greater or equal to the pivot.

code:

int partition(int arr[], int left, int right)
{
int i = left, j = right;
int tmp;
int pivot = arr[(left + right) / 2];

while (i <= j) {
while (arr[i] < pivot)
i++;
while (arr[j] > pivot)
j--;
if (i <= j) {
tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
i++;
j--;
}
};

return i;
}

void quickSort(int arr[], int left, int right) {
int index = partition(arr, left, right);
if (left < index - 1)
quickSort(arr, left, index - 1);
if (index < right)
quickSort(arr, index, right);
}

public static void main(String[] args){
...
int[] arr = new int[]{....};
.quickSort(arr,0, arr.length-1);
}

### Stretch a row if data overflows in jasper reports

It is very common that some columns of the report need to stretch to show all the content in that column. But  if you just specify the property " stretch with overflow' to that column(we called text field in jasper report world) , it will just stretch that column and won't change other columns, so the row could be ridiculous. Haven't find the solution from internet yet. So I just review the properties in iReport one by one and find two useful properties(the bold highlighted in example below) which resolve the problems.   example:
<band height="20" splitType="Stretch"> <textField isStretchWithOverflow="true" pattern="" isBlankWhenNull="true"> <reportElement stretchType="RelativeToTallestObject" mode="Opaque" x="192" y="0" width="183" height="20"/> <box leftPadding="2"> <pen lineWidth="0.25"/> …

### Live - solving the jasper report out of memory and high cpu usage problems

I still can not find the solution. So I summary all the things and tell my boss about it. If any one knows the solution, please let me know.

Symptom: 1.The JVM became Out of memory when creating big consumption report 2.Those JRTemplateElement-instances is still there occupied even if I logged out the system
Reason:         1. There is a large number of JRTemplateElement-instances cached in the memory 2.The clearobjects() method in ReportThread class has not been triggered when logging out
Action I tried:      About the Virtualizer: 1.Replacing the JRSwapFileVirtualizer with JRFileVirtualizer 2.Not use any FileVirtualizer for cache the report in the hard disk Result: The japserreport still creating the a large number of JRTemplateElement-instances in the memory     About the work around below,      I tried: item 3(in below work around list) – result: it helps to reduce  the size of the JRTemplateElement Object                Item 4,5 – result : it helps a lot to reduce the number of  JRTemplateE…