Three methods of counting and sorting

// Count_sort.cpp: defines the entry point of the console application. # Include "stdafx. H" # include <iostream> using namespace STD; const int Len = 100; Class countsort // count sorting class {public: countsort ();~ Countsort (); void fristsort (); void secondsort (); void thirdsort (); friend ostream & operator <(ostream & out, const countsort & countsort); Private: int * arr; int length ;}; countsort: countsort (): length (LEN) {arr = new int [length]; for (INT I = 0; I <length; I ++) {arr [I] = rand () % 1000 ;}} countsort ::~ Countsort () {Delete [] arr; arr = NULL;} ostream & operator <(ostream & out, const countsort & countsort) {for (INT I = 0; I <countsort. length; I ++) {cout <countsort. arr [I] <";}cout <Endl; Return out;}/* Method 1: For each element in the original array, count Array records the number of elements smaller than it. This number is the subscript of this element in the new sorted array */void countsort: fristsort () {int * COUNT = new int [length]; // for each element in the original array, the Count Array records the number of elements smaller than it. memset (count, 0, length * sizeof (INT); int * rarr = new int [length]; // rarr is the newly sorted array memset (rarr, 0, length * sizeof (INT); For (INT I = 0; I <length; I ++) {for (Int J = I + 1; j <length; j ++) {If (ARR [J] <arr [I]) {count [I] ++;} else {count [J] ++ ;}} rarr [count [I] = arr [I] ;}for (INT T = 0; t <length; t ++) {arr [T] = rarr [T];} Delete [] count; Delete [] rarr;}/* Method 2: use a temporary array to record the number of times each element appears in the original array */void countsort: secondsort () {int max = 0; // Max records the maximum element value in the array for (INT I = 0; I <length; I ++) {If (ARR [I]> MAX) {max = arr [I]; // find the maximum value} int * COUNT = new int [Max + 1]; // The Count Array records the number of occurrences of each element in the array memset (count, 0, (MAX + 1) * sizeof (INT )); int * rarr = new int [Max + 1]; // The rarr array stores the sorted element memset (rarr, 0, (MAX + 1) * sizeof (INT )); for (INT I = 0; I <length; I ++) {count [arr [I] ++; // number of record elements} For (Int J = 1; j <= max; j ++) {count [J] + = count [J-1];} For (int t = length-1; t> = 0; t --) {rarr [count [arr [T]-1] = arr [T]; count [arr [T] -- ;}for (INT z = 0; Z <length; Z ++) {arr [Z] = rarr [Z];} Delete [] count; Delete [] rarr;}/* method 3: use a temporary array to record the number of times each element appears in the original array */void countsort: thirdsort () {int max = 0; // Max records the maximum element value in the array for (INT I = 0; I <length; I ++) {If (ARR [I]> MAX) {max = arr [I]; // find the maximum value} int * COUNT = new int [Max + 1]; // The Count Array records the number of occurrences of each element in the array memset (count, 0, (MAX + 1) * sizeof (INT); For (INT I = 0; I <length; I ++) {count [arr [I] ++; // number of record elements} int z = 0; For (INT I = 0; I <= max; I ++) {While (count [I] --> 0) {arr [Z ++] = I ;}} Delete [] count ;} int _ tmain (INT argc, _ tchar * argv []) {countsort * pcountsort = new countsort (); cout <"Before sorting:" <Endl; cout <* pcountsort; // pcountsort-> fristsort (); pcountsort-> secondsort (); // pcountsort-> thirdsort (); cout <"after sorting: "<Endl; cout <* pcountsort; System (" pause "); Return 0 ;}

Vs2008 is running correctly. If you have any questions, please correct me!

Counting sorting assume that each of the N input elements is an integer between 0 and K. Here K is an integer. When k = O (N, the running time of counting sorting is O (n ).

The basic idea of counting sorting: For each input element x, determine the number of elements smaller than or equal to X. With this information, you can direct X

Place it in the output array.

Not applicable: the array contains negative values or a small number is extremely large.

Algorithm analysis:

1. the time complexity is O (n ).

2. The space complexity is O (n ).

3. Counting sorting is notIn-situ sortingAlgorithm (sorting without applying for extra space );

YesStabilitySorting Algorithm (that is, the relative order between the same keywords remains unchanged before and after sorting );

It is not a comparison-based sorting algorithm. The time complexity of the comparison sorting algorithm is O (n * logn ).