Algorithm Note _129: Count sort (Java)

Directory

1 Problem Description

2 Solutions

2.1 comparison count sort

2.2 Distribution Count sort

 1 problem description

Given a set of data, use count sorting to get this set of data from small to large sort sequences.

2 Solutions 2.1 Comparison count sort

the time complexity of the algorithm below is O (n^2), space complexity is o (n). This method is sortable for any set of data.

The specific code is as follows:

 PackageCom.liuzhen.practice; Public classMain { Public voidComparisoncountingsort (int[] A) {int[] temp =New int[A.length]; int[] Count =New int[A.length];  for(inti = 0;i < a.length-1;i++) {             for(intj = i + 1;j < a.length;j++) {                if(A[i] <A[j]) count[j]++; ElseCount[i]++; }        }         for(inti = 0;i < a.length;i++) Temp[count[i]]=A[i];  for(inti = 0;i < a.length;i++) A[i]=Temp[i]; return; }         Public Static voidMain (string[] args) {main test=NewMain (); int[] A = {2,3,1,4,6,4,3,5,3,2,5,3,5,2,3,4,5,2,54,3,21};        Test.comparisoncountingsort (A);  for(inti = 0;i < a.length;i++) System.out.print (A[i]+" "); }}

Operation Result:

2.2 Distribution count sort

the time complexity of the algorithm below is O (n), spatial complexity is o (n). The time efficiency of this method is better than that of fast sorting and merging, but this method has certain requirements for the given data, that is, the elements in the array satisfy min <= a[i] <= max, and at min ~ Max all elements between the array A have appeared in both.

The specific code is as follows:

 PackageCom.liuzhen.practice; Public classMain1 {//all of the elements in array A are greater than or equal to min, less than or equal to Max, and all min~max between them appear in array a     Public voidDistributioncountingsort (int[] A,intMinintmax) {        int[] temp =New int[A.length]; int[] D =New int[Max-min + 1];  for(inti = 0;i < a.length;i++) D[a[i]-min]++;  for(inti = 0;i < max-min;i++) D[i+ 1] = D[i + 1] +D[i];  for(inti = 0;i < a.length;i++) {            intj = A[i]-min; TEMP[D[J]-1] =A[i]; D[J]--; }         for(inti = 0;i < a.length;i++) A[i]=Temp[i]; return; }         Public Static voidMain (string[] args) {Main1 test=NewMain1 (); int[] A = {1,2,3,4,5,6,7,8,9,2,4,5,4,3,4,5,2,3,4,5,2,3,5,4,2,3,5,4,3,2,5,3,3,5}; Test.distributioncountingsort (A,1, 9);  for(inti = 0;i < a.length;i++) System.out.print (A[i]+" "); }}

Operation Result:

Resources:

1."Basic algorithm design and analysis" version 3 anany levitin Pan Yan translation

Algorithm Note _129: Count sort (Java)