C + + count sort detailed _c language

The counting sort differs from the comparison sort, is based on the counting method, and for counting, suppose each input is an integer between 0~k. For each INPUT element x, determine the number of elements less than x. If there are 17 elements less than x, then x is the 18th output position.
Counting sort involves three arrays a[0.....length-1],length the length of an array A, array B is equal to the length of array A, storing the result of the final order; c[0. K] holds the number of each element in a, K is the maximum value in array a.

int count_k (int a[],int length), which determines the length of the C array in order to determine the largest element in array A.

int count_k (int a[],int length)
{
  int j,max;
  max = a[0];

  for (j=1;j<=length-1;j++)
  {
    if (a[j]>=max)
      max = A[j];
  }

  return max;
}

Implementation of Count Sort:

void Count_sort (int a[],int b[],int k)
{
  int *c = (int *) malloc ((k+1) * sizeof (int));
  int i,j;
  for (i=0;i<=k;i++)//Initialize array C
    c[i]=0;

  for (j=0;j<=length-1;j++)//Calculate the number of elements in a
    c[a[j] = c[a[j]]+1;
  for (i=1;i<=k;i++)//To calculate the number of elements less than or equal to C[i]
    c[i] = C[i] + c[i-1];
  for (j=length-1;j>=0;j--)
  {
    int k=c[a[j]]-1;
    B[K] = a[j];
    C[A[J]] = C[a[j]]-1;

  }

  Free (C);

}

Count_sort (A,B,K);

K=5

for (j=0;j<=length-1;j++)//Calculate the number of elements in a    
c[a[j] = c[a[j]]+1;

Represents an array of 2 0, 0 1, 2 2, 3 3, 0 4, and 5

  for (i=1;i<=k;i++)//To calculate the number of elements less than or equal to C[i]    
c[i] = C[i] + c[i-1];

A number less than or equal to 0 has two, a number less than or equal to 1 has two, 2 is less than 4, 3 is less than or equal to 7, 4 is less than or equal to 7, and 5

for (j=length-1;j>=0;j--)
  {
    int k=c[a[j]]-1;
    B[K] = a[j];
    C[A[J]] = C[a[j]]-1;

  }

For loop analysis is as follows

j=7; a[j]=a[7]=3; c[a[j]]=c[3]=7; c[a[j]]-1=6; b[c[a[j]]-1]=b[6]=a[j]=3; C[a[j]]=c[a[j]]-1=6

j=6; a[j]=a[6]=0; c[a[j]]=c[0]=2; C[a[j]]-1=1; b[c[a[j]]-1]=b[1]=a[j]=0; C[a[j]]=c[a[j]]-1=1

j=5; a[j]=a[5]=3; c[a[j]]=c[3]=6; c[a[j]]-1=5; b[c[a[j]]-1]=b[5]=a[j]=3; c[a[j]]=c[a[j]]-1=5;

j=4; a[j]=a[4]=2; c[a[j]]=c[2]=4; c[a[j]]-1=3; b[c[a[j]]-1]=b[3]=a[j]=2; c[a[j]]=c[a[j]]-1=3;

j=3; a[j]=a[3]=0; C[a[j]]=c[0]=1; c[a[j]]-1=0; b[c[a[j]]-1]=b[0]=a[j]=0; c[a[j]]=c[a[j]]-1=0;

j=2; a[j]=a[2]=3; c[a[j]]=c[3]=5; c[a[j]]-1=4; b[c[a[j]]-1]=b[4]=a[j]=3; c[a[j]]=c[a[j]]-1=4;

J=1; a[j]=a[1]=5; c[a[j]]=c[5]=8; c[a[j]]-1=7; b[c[a[j]]-1]=b[7]=a[j]=5; c[a[j]]=c[a[j]]-1=7;

j=0; a[j]=a[0]=2; c[a[j]]=c[2]=3; c[a[j]]-1=2; b[c[a[j]]-1]=b[2]=a[j]=2; c[a[j]]=c[a[j]]-1=2;

Count sort of last run screenshot

Count Sort analysis: j=length-1;j>=0;j– here for the reverse, is to ensure the stability of the order, this in the cardinal order has an important role.