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());
    }
}