我们假定输入是个数组A[0...n]，length[A]=n，另外，我们还需要两个数组：存放结果的B[1..n]，以及提供临时存储的C[0...k].
COUNT_SORT(A,B,k)
1. for i=0 to k
2. do C[i]=0
3. for j=1 to length[A]
4, do C[A[ j ]]=C[A[ j ]] + 1
5, for i =1to k
6, do C[i]=C[i] + C[i-1]
7, for j = length[A] downto 1
8, do B[C[A[ j ]]] =A[j]
9, C[A[ j ]] = C[A[ j ]] -1