Computerscience analysis of sorting algorithms show that, on average, no generic sorting algorithm can scale faster than O(nlogn) (see the sidebar Orders of Magnitude). However, many applications don't require a "general" sort. You often have additional information that can help you to improve the speed of a particular sort.
Orders of MagnitudeWhen discussing the time taken for particular algorithms to execute, it is important to know not just how long the algorithm takes for a particular dataset, but also how long it takes for differentsized datasets, i.e., how it scales. For applications, the problems of handling 10 objects and handling 10 million objects are often completely different problems, not just differentsized versions of the same problem. One common way to indicate the behavior of algorithms across different scales of datasets is to describe the algorithm's scaling characteristics by the dominant numerical function relating to the scaling behavior. The notation used is "O(function)," where function is replaced by the dominant numerical scaling function. It is common to use the letter "n" to indicate the number of data items being considered in the function. For example, O(n) indicates that the algorithm under consideration increases in time linearly with the size of the dataset. O(n^{2}) indicates that the time taken increases according to the square of the size of the dataset. These orders of magnitude do not indicate absolute times taken by the algorithm. Instead, they indicate how much longer the algorithm takes when the dataset increases in size. If an O(n) algorithm takes 200 seconds for n=10, it will take about 2000 seconds for n=100, i.e., a tenfold increase in the dataset size implies a tenfold increase in the amount of time taken by the algorithm. An O(n^{2}) algorithm might take 5 seconds for n=10, and about 500 seconds for n=100, i.e., a tenfold increase in the dataset size implies a hundredfold increase in the time taken by the algorithm. Note that the scaled times are approximate; orderofmagnitude statistics include only the dominant scaling function, and there may be other, smaller terms that adjust the actual time taken. The order of magnitude does not indicate the relative speeds of two different algorithms for any specific dataset size. Instead, orderofmagnitude statistics indicate how expensive one particular algorithm may be as your dataset grows. In the examples of the last paragraph, the time taken for the second O(n^{2}) algorithm increases much faster than the first O(n) algorithm, but the O(n^{2}) algorithm is still faster at n=100. However, by n=1000, it would be the slower of the two algorithms (50,000 seconds, compared to 20,000 seconds for the O(n) algorithm). To take a concrete example, hash tables have an O(1) order of magnitude for accessing elements. This means that the time taken to access elements in a hash table is mostly independent of the size of the hash table. Accessing elements in an array by linearly searching through that array takes O(n). In absolute times, it might be quicker to execute the linear array search on a small array than to access from a hash table. But as the number of elements becomes larger, at some point the hash table will always become quicker. 
For example, if you have 1000 items to sort and each item can be given a unique ordering value that corresponds to a unique integer between 1 and 1000, the best sort is simply to slot the items directly into their correct locations in an array. No comparisons between the items are necessary.
Of course, typically you can't map your elements so neatly. But if you can map items to integer keys that are more or less evenly distributed, you can still take advantage of improved sorting characteristics. Bear in mind that an array of partially sorted items can be sorted faster than a typical unsorted array.
When you can guess the approximate final position of the items in the collection to be sorted, you can use this knowledge to improve sorting speed. You should specifically look out for sorts where:
Items can be given an ordering value that can be mapped to integer keys.
The distribution of the keys is regular, or any one of the following is true:
The distribution of the keys is fairly even, so that when mapped into array indexes, ordering is approximately kept.
The keys have evenly distributed clusters.
The distribution of the keys has a mapping into one of these other distributions.
The distribution of the keys is fairly critical. A regular distribution allows them to be mapped straightforwardly into array indexes. An uneven distribution is difficult to map. But if you have an uneven distribution and can specify a mapping that allows you to flatten out the keys in some way, it may still be possible to apply this methodology. For example, if you know that your keys will have a normal (bellcurve) distribution, you can apply an inverse bellcurve function to the keys to flatten them out to array indexes.
For this technique to work, the mapped keys do not need to be unique. Several keys or groups of keys can map to the same value or values. Indeed, it is quite difficult to make the index mapping unique. You need to be aware of this and handle the resulting collisions. Normally, you can map clusters of keys into subsections of the sorted array. These subsections are probably not internally sorted, but they may be correctly sorted against each other (i.e., all elements in subsection 1 are ordered below all elements in subsection 2, all elements in subsection 2 are ordered below all elements in subsection 3, etc.). This way, the problem has been modified to sort multiple smaller subsections (which is faster than sorting the whole array), and hence the array is sorted more quickly.
Note that Object.hashCode( ) provides a mechanism for generating an integer for any object. However, the resulting hash code is not guaranteed to be evenly distributed or even unique, nor is it at all guaranteed to be consistent across different VMs or even over multiple runs of one VM. Consequently, the hash code is of little use for any kind of mapping.
KarlDietrich Neubert^{[4]} gives a detailed implementation of this approach, where the algorithm provides O(n) sorting behavior and also minimizes the extra memory needed to manage the sort.
^{[4]} Algorithm Alley, Dr. Dobb's Journal, February 1998.

I include here a Java implementation of Neubert's algorithm with comments in the code. I have applied the implementation to an array of objects that have an integer field to specify the sort order, but of course the algorithm can be easily generalized to other cases where object ordering can be mapped to numbers. For a more detailed discussion of the algorithm, refer to Neubert's article. The implementation given here sorts significantly faster than either the sort in java.util.Arrays (in Java 2) or a handcrafted quicksort (the most optimized final version with no casts from the first section of this chapter); see Table 93. Note also that this sort is O(n) and thus increases linearly in time, whereas the other sorts are O(nlogn) and so have a superlinear speedup. Notice how the interpreted mode Flashsort time approaches the timings of the compiled JVMs and is actually faster than the pure JIT of 1.2.2 running Arrays.sort( ). That's a significant speedup.
1.2.2 
1.3.1_02 
1.3.1_02server 
1.4.0 
1.4.0server 
1.4.0Xint 


Neubert's Flashsort 
100% 
94% 
155% 
74% 
79% 
294% 
Handcrafted quicksort 
250% 
178% 
130% 
149% 
132% 
921% 
Arrays.sort( ) 
476% 
181% 
183% 
174% 
164% 
906% 
Note that the sort at the end of the Neubert algorithm is an insertion sort running over the entire array. Insertion sorts provide better performance than quicksorts for partially ordered arrays. This final insertion sort ensures that keys incorrectly classified by the group distribution end up in the right location:
public interface FlashSortable{
public int sortOrder( );
}
public static void flashsort(FlashSortable[ ] arr)
{
//Number of groups into which the elements are classified
//Neubert suggests 0.2 to 0.5 times the number of elements in the array.
int num_groups = (int) (0.4 * arr.length);
//Count the number of elements in each group
int[ ] groups = new int[num_groups];
flashsort(arr, num_groups, groups);
}
public static void flashsort(FlashSortable[ ] arr, int num_groups, int[ ] groups)
{
//First get the minimum and maximum values
int min = arr[0].sortOrder( );
int max_idx = 0;
int i;
for (i = arr.length1; i > 0; i)
{
if (arr[i].sortOrder( ) < min)
min = arr[i].sortOrder( );
if (arr[i].sortOrder( ) > arr[max_idx].sortOrder( ))
max_idx = i;
}
//If they are the same, all elements are identical
//so the array is already sorted.
if (min = = arr[max_idx].sortOrder( ))
return;
//Count the number of elements in each group.
//Take care to handle possible integer overflow by
//casting to larger datatypes where this might occur.
double scaling_constant = (num_groups  1) /
( ((double) arr[max_idx].sortOrder( ))  min);
int group;
for (i = arr.length1; i >= 0; i)
{
group = (int) (scaling_constant * (((long) arr[i].sortOrder( ))  min));
groups[group]++;
}
//Set the groups to point to the indexes in the array
//that are the last index for each group.
groups[0];
for (i = 1; i < groups.length; i++)
{
groups[i] += groups[i1];
}
//Put the biggest element at index 0 so that the swapping
//algorithm below starts on the largest element & max group.
FlashSortable old_value = arr[max_idx];
arr[max_idx] = arr[0];
arr[0] = old_value;
//start with element at index 0
int idx = 0;
//and the maximum group
group = num_groups  1;
//Start moving elements into their groups.
//We need to make 'arr.length' moves at most,
//but if we have one move left in the outer loop
//then the remaining element is already in the right place,
//so we need test for only 'arr.length1' moves.
int number_of_moves_left = arr.length  1;
FlashSortable new_value;
while(number_of_moves_left > 0)
{
//When the first group fills up, we start scanning
//for elements left in the wrong groups, and move them.
//Note that we scan through the whole object array only once.
while(idx > groups[group])
{
idx++;
group = (int) (scaling_constant * (((long) arr[idx].sortOrder( ))  min));
}
new_value = arr[idx];
//We run this loop until the first group fills up.
//Then we run the previous scan loop to get back into this loop.
while( idx != (groups[group]+1) )
{
group = (int) (scaling_constant * (((long) new_value.sortOrder( ))  min));
old_value = arr[groups[group]];
arr[groups[group]] = new_value;
new_value = old_value;
groups[group]; //decrement the pointer to the next index
number_of_moves_left;
}
}
//Now we have our partially ordered array,
//we do an insertion sort to order the remainder.
for (i = arr.length  3; i >= 0; i)
{
if (arr[i+1].sortOrder( ) < arr[i].sortOrder( ))
{
old_value = arr[i];
idx = i;
while(arr[idx+1].sortOrder( ) < old_value.sortOrder( ))
{
arr[idx] = arr[idx+1];
idx++;
}
arr[idx] = old_value;
}
}
}