cs61b homework10

part1：countingsort的算法感觉还是蛮绕的，贼烧脑，注意是对16进制的每一位进行排序，取得相应位数值算法可以写成(keys[i]%(whichDigit+1))/whichDigit;

code：

public static int[] countingSort(int[] keys, int whichDigit) {
        if(whichDigit>7||whichDigit<0)
            return null;
        int sort[]=new int[keys.length];
        double num=Math.pow((double)16, (double)(whichDigit+1));
        double num3=Math.pow((double)16, (double)(whichDigit));
        int num2=(int)num;
        int num4=(int)num3;
        for(int i=0;i<keys.length;i++){
            sort[i]=(keys[i]%num2)/num4;
        }
        int count[]=new int[16];
        for(int i=0;i<sort.length;i++){
        count[sort[i]]++;    
        }
        int total=0;
        for(int i=0;i<count.length;i++){
            int c=count[i];
            count[i]=total;
            total+=c;
        }
        int y[]=new int[keys.length];
        for(int i=0;i<keys.length;i++){
            y[count[sort[i]]]=keys[i];
            count[sort[i]]++;
        }
        return y;
        
        
      }

part2：radixsort在coutingsort基础上叠加7次，到int的取值范围即可。

code：

public static int[] radixSort(int[] keys) {
          int[]radixsort=new int[keys.length];
          for(int i=0;i<keys.length;i++)
              radixsort[i]=keys[i];
        for(int i=0;i<8;i++){
            radixsort=countingSort(radixsort,i);
        }
        return radixsort;
      }

运行结果：

keys are [ 60013879 11111119 2c735010 2c732010 7fffffff 4001387c 10111119 529a7385 1e635010 28905879 11119 0 7c725010 1e630010 111111e5 61feed0c 3bba7387 52953fdb 40013879 ]
keys are [ 0 11119 7fffffff 10111119 11111119 111111e5 1e630010 1e635010 28905879 2c732010 2c735010 3bba7387 40013879 4001387c 52953fdb 529a7385 60013879 61feed0c 7c725010 ]