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Java 5 Update 5 (1.5.0_05) is the last release of Java to work on Windows 95 (with Internet Explorer 5.5 installed) and Windows NT 4.0. [36] Java 5 was first available on Apple Mac OS X 10.4 (Tiger) [37] and was the default version of Java installed on Apple Mac OS X 10.5 (Leopard). Public support and security updates for Java 1.5 ended in ...
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The complexity of the cocktail shaker sort in big O notation is () for both the worst case and the average case, but it becomes closer to () if the list is mostly ordered before applying the sorting algorithm. For example, if every element is at a position that differs by at most k (k ≥ 1) from the position it is going to end up in, the ...
1 < 5, so 1 is not added to the sub-list. 4 < 5, so 4 is not added to the sub-list. 2 < 5, so 2 is not added to the sub-list. 0 < 5, so 0 is not added to the sub-list. 9 > 5, so 9 is added to the sub-list and removed from the original list. Step 4: Now compare 9 with the remaining elements in the original list until there is a number greater than 9
The next pass, 3-sorting, performs insertion sort on the three subarrays (a 1, a 4, a 7, a 10), (a 2, a 5, a 8, a 11), (a 3, a 6, a 9, a 12). The last pass, 1-sorting, is an ordinary insertion sort of the entire array (a 1,..., a 12). As the example illustrates, the subarrays that Shellsort operates on are initially short; later they are longer ...
Natural sort order has been promoted as being more human-friendly ("natural") than machine-oriented, pure alphabetical sort order. [ 1 ] For example, in alphabetical sorting, "z11" would be sorted before "z2" because the "1" in the first string is sorted as smaller than "2", while in natural sorting "z2" is sorted before "z11" because "2" is ...
A bidirectional variant of selection sort (called double selection sort or sometimes cocktail sort due to its similarity to cocktail shaker sort) finds both the minimum and maximum values in the list in every pass. This requires three comparisons per two items (a pair of elements is compared, then the greater is compared to the maximum and the ...
def cycle_sort (array)-> int: """Sort an array in place and return the number of writes.""" writes = 0 # Loop through the array to find cycles to rotate. # Note that the last item will already be sorted after the first n-1 cycles. for cycle_start in range (0, len (array)-1): item = array [cycle_start] # Find where to put the item. pos = cycle_start for i in range (cycle_start + 1, len (array ...