Question

有没有人知道如何从这个伪代码中实现Java中的Sum-of-Subsets问题？

w = an array of positive integers sorted in non-decreasing order.
W = the target sum value
include = an array or arraylist of the solutions who's weight adds up to W. After the print statement, this array can be deleted so that it can store the next solution.
weight = weight of elements in the include array.
total = weight of the remaining elements not in the include array.

public static void sum_of_subsets(index i, int weight, int total)
{
     if(promising(i))
     {
          if(weight == W)
          {
               System.out.print(include[1] through include[i]);
          }
          else
          {
               include[i + 1] = "yes";     //Include w[i + 1]
               sum_of)subsets(i + 1, weight + w[i + 1], total - w[i + 1]);
               include[i + 1] = "no";      //Do not include w[i + 1]
               sum_of_subsets(i + 1, weight, total - w[i + 1]);
          }
     }
}

public static boolean promising(index i);
{
     return (weight + total >= W) && (weight == W || weight + w[i + 1] <= W);
}

这真让我感到困惑，所以如果你能添加评论会很棒!!!

Answer 1

Algorithms Coded in Java

首先，该算法会删除所有大于总和的数字。

然后，对于小于总和的最大数字，它检查列表中是否有任何数字可以添加到自身以获得总和。一旦我们找到一对，或者总和大于所需的总和，我们就可以打破，因为列表已经排序。然后我们考虑第二大数字，看看我们是否可以与之配对，依此类推。

   /**
    * This will find how many pairs of numbers in the given array sum
    * up to the given number.
    *
    * @param array - array of integers
    * @param sum - The sum
    * @return int - number of pairs.
    */
public static int sumOfSubset(int[] array, int sum)
{
        // This has a complexity of O ( n lg n )
        Arrays.sort(array);

        int pairCount = 0;
        int leftIndex = 0;
        int rightIndex = array.length - 1;

        // The portion below has a complextiy of
        //  O ( n ) in the worst case.
        while (array[rightIndex] > sum + array[0])
        {
            rightIndex--;    
        }

        while (leftIndex < rightIndex)
        {
            if (array[leftIndex] + array[rightIndex] == sum)
            {
                pairCount++;
                leftIndex++;
                rightIndex--;
            }
            else if(array[leftIndex] + array[rightIndex]  < sum)
            {
                leftIndex++;
            }
            else
            {
                rightIndex--;   
            }
        }

        return pairCount;
}

上面的算法不会返回对，但这很容易添加。

Answer 2

该程序从一组数字中找到确切的对以形成所需的总和，程序最后还返回唯一的对。如果没有找到确切的子集，程序还返回最近/最近的子集以形成所需的总和。您可以按原样运行程序以查看演示，然后根据需要进行修改。我在这里应用的逻辑是基于给定集合中的所有数字组合来获得所需的总和，您可以参考内联注释以获取更多信息

package com.test;

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

/**
 * 
 * @author ziya sayed
 * @email : sayedziya@gmail.com
 */
public class SumOfSubsets{

//  private static int[] numbers= {1,1,2,2,3,4};//set of numbers
    private static int[] numbers= {18,17,1};//set of numbers
//  private static final int SUM = 4;//desired sum
    private static final int SUM = 20;//desired sum

    public static void main(String[] args) {

        String binaryCount="";
        int[] nos=new int[numbers.length];
        //input set, and setting binary counter
        System.out.print("Input set numbers are : ");
        for (int no : numbers) {            
            if (no<=SUM) {
                System.out.print(no+" ");                                       
                nos[binaryCount.length()]=no;
                binaryCount+=1; //can we use sring builder or string.format
            }       

        }
        System.out.println();
        Arrays.sort(nos);//sort asc

        int totalNos = binaryCount.length();
        String subset="";   //chosen subset 
        int subsetSum=0;//to temp hold sum of chosen subset every iteration
        String nearestSubset="";//chosen closest subset if no exact subset
        int nearestSubsetSum=0;//to hold sum of chosen closest subset

        Set<String> rs = new HashSet<String>();//to hold result, it will also avoide duplicate pairs
        for (int i = Integer.parseInt(binaryCount, 2) ; i >0;i-- ) {//for all sum combinations
        //  System.out.println(i);
            binaryCount=String.format("%1$#" + totalNos + "s", Integer.toBinaryString(i)).replace(" ","0");//pad 0 to left if number is less than 6 digit binary for proper combinations

            subset="";
            subsetSum=0;

            for (int j=0 ;j<totalNos; j++) {//for active combinations sum

                if (binaryCount.charAt(j)=='1') {                   
                    subset+=nos[j]+" ";
                    subsetSum+=nos[j];
                }
            }
            if (subsetSum == SUM) {
            //  System.out.println(subset);//we can exit here if we need only one set
                rs.add(subset);
            }
            else{//use this for subset of numbers with nearest to desired sum
                if (subsetSum < SUM  && subsetSum > nearestSubsetSum && rs.isEmpty()) {
                    nearestSubsetSum = subsetSum;
                    nearestSubset = subset;

                }
            }

        }

        if (rs.isEmpty()) {
                System.out.println("Nearest Subset of "+SUM);
                System.out.println(nearestSubset);
        }
        else{
            System.out.println("Exact Subset of "+SUM);
            System.out.println(rs);//unique sub sets to remove duplicate pairs
        }

    }


}

如何在Java中实现子集和的问题

2 个答案: