Introduction to algorithms learning notes-counting and sorting algorithms

/*** Counting sorting assume that each of the N input elements is an integer between 0 and K. Here K is an integer. * The basic idea is: For each input element, determine the number of elements smaller than X, so that X can be directly placed in its position in the final output array. *AlgorithmThe complexity is O (n + k). When k = O (n), the running time is O (n) * another characteristic: the counting sorting is stable, the relative order of elements with the same value in the output array is the same as the order of the elements in the input array */public class countingsort {/*** @ Param A // stores the original array to be sorted * @ Param B // The array that stores the sorting result * @ Param K // each element in a is between 0 and K */static void countingsort (int A [], int B [], int K) {// The C array is a temporary storage area. c stores the number of times each element appears in a from 0 to K, therefore, the length of C is k + 1. Int C [] = new int [k + 1]; // The number of elements in C [I] That store equal to I (0 = <I <= K) for (INT I = 0; I <. length; I ++) C [A [I] ++; // C [I] stores the number of elements smaller than or equal to I for (INT I = 1; I <= K; I ++) C [I] = C [I] + C [I-1]; // Since C [I] stores less than or equal to the number of I elements, therefore, I should be placed in the C [I] position in array B (because the array starts from 0, so 1 is required) for (INT r =. length-1; r> = 0; r --) {int I = A [R]; B [C [I]-1] = I; // every time C [I] minus 1, it can ensure that the next element is the same as I, which falls in the previous position C [I] --;} of B [C [I]-1]. // test ..... public static void main (string [] ARGs) {int A [] = {2, 5, 3, 0, 2, 3, 0}; int B [] = new int [8]; countingsort (A, B, 5); For (INT I = 0; I <B. length; I ++) system. out. println (B [I]);}