Coursera Algorithms week2 基础排序 练习测验: Permutation

题目原文：

Given two integer arrays of size n , design a subquadratic algorithm to determine whether one is a permutation of the other. That is, do they contain exactly the same entries but, possibly, in a different order.

本质上就是求两个数组排序后是否相等，鉴于本节课学的是选择、插入、希尔排序，就选个最不熟悉的希尔排序来实现吧

 import java.util.Arrays;

 public class PermutationArrays {
     private int[] a;
     private int[] b;
     private int n;
     PermutationArrays(int n, int[] a, int[] b){
         this.a = a;
         this.b = b;
         this.n = n;
     }
     private boolean less(int v, int w){
         return v < w;
     }
     private void exch(int[] a, int i, int j){
         int t = a[i];
         a[i] = a[j];
         a[j] = t;
     }
     private void sortByShell(int n, int[] a){
         int h = 1;
         while(h < n/3){
             h = 3*h + 1;
         }
         while(h>=1){
             for(int i = h; i<n;i++){
                 for(int j = i; j>=h && less(a[j],a[j-h]);j=j-h){
                     exch(a,j,j-h);
                 }
             }
             h = h/3;
         }
     }
     public boolean isPermutation(){
         sortByShell(n,a);
         System.out.println(Arrays.toString(a));
         sortByShell(n,b);
         System.out.println(Arrays.toString(b));
         for(int i=0;i<n;i++){
             if(a[i] != b[i])
                 return false;
         }
         return true;
     }
     public static void main(String[] args){
         int[] a = {1,2,4,5,6,11,9,7,8,0};
         int[] b = {0,9,8,7,6,5,4,3,2,1};
         PermutationArrays pa = new PermutationArrays(10,a,b);
         System.out.println(pa.isPermutation());
     }
 }