There are different algorithms for sorting an array of strings. Time-critical client applications often use the algorithm called Quicksort, which is very efficient for such applications. This two-part article series takes a close look at writing and using such a function.
Quicksort - Quicksort by Observation (Page 4 of 5 )
Let us sort the list above by observation, before we see how to program it on the computer.
The list is:
Q W E R T Y U I O P
Let us make R the pivot. We can equally make Q, whose rightful spot is position 4, the pivot. We simple have to choose something for the central position. So by choosing R and exchanging the positions of R and Y, we have:
Q W E Y T R5 U I O P
When any element obtains its rightful position, we shall indicate that with a superscript as in the case of R above.
We now have to see that all the elements to the left of R are less than R, and that all the elements to the right of R are greater than R, by exchanging pairs. The following pairs can be exchanged: W and P, and Y and O. Note that you can still exchange W and O, and Y and P. Exchanging W for P and Y for O, we have
[Q P E O T] (R5) [U I Y W] ------------result I
We will indicate the sub-lists with square brackets, and the pivot with normal brackets.
If you chose O as the pivot for the left sub-list and repeated the first and second steps in the algorithm, you would have something like this:
[I E] (O2) [P Q] ------------result IIa
If we do a similar thing to the right sub-list above we shall have something like:
[T] (U7) [Y W]
Now when a non-pivot element appears alone in a sub-list, we know that the element has arrived at its rightful position. So we have
[T6] (U7) [Y W] ------------result IIb
The state at which we have sorted the whole list by observation, so far, is
Note that some elements have taken their final (rightful) positions. Any element that is a pivot, obtains its rightful position. Any element that is alone in a sub-list also obtains its rightful position.
You simply have to continue by repeating the first and second steps of the algorithm on the sub-lists. Let us consider the first sub-list, [I E] of result III.
There are only 2 elements here. Let us choose E as the pivot. Putting E in its rightful position by exchanging it with “I”, we have:
(E0) [I]
Now, the position number of E in the main list is zero; remember that array counting begins from zero. There is only one sub-list, the right sub-list. It has just one element, so the element has arrived at its proper position, which is position 1. The former sub-list is:
(E0) [I1] --------------result IVa
You can carry out a similar analysis for the other sub-lists and come out with the complete sorted list of
JavaScript 
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QuickSort in Action
