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循环分析如下
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;
