Coursera Algorithms week3 快速排序 练习测验: Nuts and bolts

Posted 2020-09-29 evasean

题目原文：

Nuts and bolts. A disorganized carpenter has a mixed pile of n nuts and n bolts. The goal is to find the corresponding pairs of nuts and bolts. Each nut fits exactly one bolt and each bolt fits exactly one nut. By fitting a nut and a bolt together, the carpenter can see which one is bigger (but the carpenter cannot compare two nuts or two bolts directly). Design an algorithm for the problem that uses nlogn compares (probabilistically). 

分析：

题意是有一堆螺帽和螺钉，分别为n个，每个螺帽只可能和一个螺钉配对，目标是找出配对的螺帽和螺钉。螺帽和螺钉的是否配对只能通过螺帽和螺钉比较，不能通过两个螺帽或两个螺钉的比较来判断。比较次数要求限制在nlogn次

设计过程中思考了如下几个问题： 

1. 螺帽和螺钉的配对怎么判断？

　　-螺帽和螺钉分别设计成不同的对象，每个对象都有个size属性，通过判断不同对象的size是否相等来判断是否配对

2. 为什么不能通过把螺帽和螺钉分别排序，然后对应位置一一配对的方式进行设计？

　　-假如那堆螺帽和螺钉中分别有落单的不能配对的，这种排序后靠位置来匹配的配对方式就明显不合适了，也就是说这种做法鲁棒性太差

3. 既然不能分别排序，那采用把螺帽和螺钉混在一起排序的方式如何？

　　-恩，貌似可行，但遇到螺帽和螺钉中分别有落单的不能配对的情况，我怎么判断某个位置i处元素是与i-1处的元素配对？还是与i+1处的元素配对？还是i处元素落单呢？

综上几个问题考虑之后，决定如下设计：

a. 现将螺帽进行快速排序，复杂度nlogn

b. 逐个遍历螺钉组中的每个螺钉，在已排序的螺帽中，采用二分查找的方法查找其配对的螺帽。比较次数nlogn，满足题目要求

代码如下

 1 package week3;
 2 /**
 3  * 螺帽和螺钉共有父类
 4  * @author evasean www.cnblogs.com/evasean/
 5  */
 6 public class NBParent {
 7     public NBParent(int size){
 8         this.size = size;
 9     }
10     private int size;
11     public int getSize() {
12         return size;
13     }
14     public void setSize(int size) {
15         this.size = size;
16     }
17 }

 1 package week3;
 2 /**
 3  * 螺帽类
 4  * @author evasean www.cnblogs.com/evasean/
 5  */
 6 public class Nut extends NBParent{
 7     public Nut(int size){
 8         super(size);
 9     }
10 }

 1 package week3;
 2 /**
 3  * 螺钉类
 4  * @author evasean www.cnblogs.com/evasean/
 5  */
 6 public class Bolt extends NBParent{
 7     public Bolt(int size){
 8         super(size);
 9     }
10 }

  1 package week3;
  2 
  3 import java.util.HashMap;
  4 import java.util.Iterator;
  5 import java.util.Map;
  6 import java.util.Map.Entry;
  7 import edu.princeton.cs.algs4.StdRandom;
  8 /**
  9  * 螺帽类
 10  * @author evasean www.cnblogs.com/evasean/
 11  */
 12 public class NutsAndBolts {
 13     Map<Nut, Bolt> pairs = new HashMap<Nut, Bolt>(); // 存储配对的螺帽和螺丝对
 14     Nut[] nuts;
 15     Bolt[] bolts;
 16     int n;
 17 
 18     public NutsAndBolts(Nut[] nuts, Bolt[] bolts, int n) {
 19         this.nuts = nuts;
 20         this.bolts = bolts;
 21         this.n = n;
 22     }
 23 
 24     private int compare(NBParent v, NBParent w) {
 25         int vsize = v.getSize();
 26         int wsize = w.getSize();
 27         if (vsize == wsize) return 0;
 28         else if (vsize > wsize) return 1;
 29         else return -1;
 30     }
 31     private void exch(NBParent[] nb, int i, int j){
 32         NBParent t = nb[i];
 33         nb[i]=nb[j];
 34         nb[j]=t;
 35     }
 36     
 37     public Map<Nut, Bolt> findPairs() {
 38         sort(bolts,0,n-1); //先对bolts进行快速排序
 39         for(int i = 0; i<n;i++){ //遍历nuts，并在bolts中寻找其成对的bolt
 40             Nut nut = nuts[i];
 41             Bolt bolt= findBolt(nut); 
 42             if(bolt != null)
 43                 pairs.put(nut, bolt);
 44         }
 45         return pairs;
 46     }
 47     private Bolt findBolt(Nut nut){ //在排好序的bolts中二分查找nut
 48         int lo = 0; 
 49         int hi = n-1;
 50         while(lo<=hi){
 51             int mid = lo+(hi-lo)/2;
 52             int cr = compare(bolts[mid],nut);
 53             if(cr<0) lo = mid+1;
 54             else if(cr>0) hi = mid-1;
 55             else return bolts[mid];
 56         }
 57         return null;
 58     }
 59     private void sort(NBParent[] nb, int lo, int hi){
 60         if(hi<=lo) return;
 61         int j = partition(nb,lo,hi);
 62         sort(nb,lo,j-1);
 63         sort(nb,j+1,hi);
 64     }
 65     
 66     private int partition(NBParent[] nb, int lo, int hi){
 67         int i = lo;
 68         int j = hi+1;
 69         NBParent v = nb[lo];
 70         while(true){
 71             while(compare(nb[++i],v)<0) if(i==hi) break;
 72             while(compare(nb[--j],v)>0) if(j==lo) break;
 73             if(i>=j) break;
 74             exch(nb,i,j);
 75         }
 76         exch(nb,lo,j);
 77         return j;
 78     }
 79 
 80     public static void main(String[] args) {
 81         int n = 10;
 82         Nut[] nuts = new Nut[n];
 83         Bolt[] bolts = new Bolt[n];
 84         for (int i = 0; i < n-1; i++) {
 85             Nut nut = new Nut(i + 1);
 86             nuts[i] = nut;
 87             Bolt bolt = new Bolt(i + 2);
 88             bolts[i] = bolt;
 89         }
 90         //故意做一对不一样的
 91         nuts[n-1] = new Nut(13);//nuts的size分别为{1,2,3,4,5,6,7,8,9,13}
 92         bolts[n-1] = new Bolt(1);//bolts的size分别是{2,3,4,5,6,7,8,9,10,1}
 93         StdRandom.shuffle(nuts);
 94         StdRandom.shuffle(bolts);
 95         NutsAndBolts nb = new NutsAndBolts(nuts, bolts, n);
 96         Map<Nut, Bolt> pairs = nb.findPairs();
 97         Iterator<Entry<Nut, Bolt>> iter = pairs.entrySet().iterator();
 98         while(iter.hasNext()){
 99             Entry<Nut, Bolt> e = iter.next();
100             Nut nut = e.getKey();
101             Bolt bolt = e.getValue();
102             System.out.print("<"+nut.getSize()+","+bolt.getSize()+">,");
103         }
104         System.out.println();
105     }
106 }