我有一组这样的整数

{1,4,5,2,7,8,-3,-5,-6,9,3,-7,-1,5,6} 

该集合可以包含任意数量的项目，因为输入是从用户那里获取的，我需要从该集合中找到所有可能的子集，其总和等于零，例如在这种情况下，在上面的集合中，子集将是

{(1,2,-3)}

{(1,-1)}

{(3,-3)}

{(5,-5)}

等等等等

我试过这段代码，但是当我将 target 设置为零时它没有返回答案。

import java.util.ArrayList;
import java.util.Arrays;

class SumSet {

    static void sum_up_recursive(ArrayList<Integer> numbers, int target,
                                 ArrayList <Integer> partial)
    {
        int s=0;
        for (int x: partial) s += x;
        if (s == target)
            System.out.println("sum("+Arrays.toString(partial.toArray())+")="+target);

        if (s >= target)
            return;

        for(int i=0;i<numbers.size();i++) {
            ArrayList<Integer> remaining = new ArrayList<Integer>();
            int n = numbers.get(i);
            for (int j=i+1; j<numbers.size();j++) remaining.add(numbers.get(j));
            ArrayList<Integer> partial_rec = new ArrayList<Integer>(partial);
            partial_rec.add(n);
            sum_up_recursive(remaining,target,partial_rec);
        }
    }

    static void sum_up(ArrayList<Integer> numbers, int target) 
    {
        sum_up_recursive(numbers,target,new ArrayList<Integer>());
    }

    public static void main(String args[]) {
        Integer[] numbers = {3,4,6,4,5,2,6};
        int target = 9;
        sum_up(new ArrayList<Integer>(Arrays.asList(numbers)),target);
    }
}

最佳答案

这是一个解决方案。

我首先解决第一个子问题:我假设所有数字和目标都是正数，然后我解决真正的问题。

为了实现这一点，我基本上将问题分解为子问题。

让我们用一个例子来说明:

数字:1,3,8,2,7 目标:10

首先:对列表进行排序: 数字:8,7,3,2,1 目标:10 然后递归地找到以下问题的解决方案:

数字:7,3,2,1 目标:10-8 = 2

数字:3,2,1 目标:10-7=3

数字:2,1 目标:10-3=2

数字:1 目标:10-1=9

前面放大数的目的是为了快速排除包含这个数的解(因为总和很快就超过了目标)。

下面是解决这个子问题的注释代码:

import java.util.ArrayList;
import java.util.List;

public class Problem {

    /*
     * Used at the end to recompose the solutions.
     * This value is null for the root problem.
     */
    private Integer nodeValue;

    //The POSITIVE target sum
    private int target;

    //List of POSITIVE numbers, supposed to be sorted
    private List<Integer> numbers;

    private List<Problem> listSubProblems;

    /*
     * Link to the parent problem.
     * Useful at the end to generate the results.
     */
    private Problem parentProblem;

    public Problem(int target, List<Integer> numbers, Integer nodeValue, Problem parentProblem){
        this.target=target;
        this.numbers=numbers;
        this.nodeValue=nodeValue;
        this.parentProblem=parentProblem;
        this.listSubProblems =new ArrayList<Problem>();
    }

    public void solve(){
        buildSubProblems();
        for(Problem problem : listSubProblems){
            problem.solve();
        }
    }

    /**
     * Builds a List of sub problems.
     * For example, if {@link #numbers} contains 9 8 5 3, with target 10
     * this method will return the following sub problems:
     *
     * <table>
     *     <tr>
     *         <td>nodeValue</td><td>target</td><td>numbers</td>
     *     </tr>
     *     <tr>
     *         <td>9</td><td>10-9=1</td><numbers>8 5 3</numbers>
     *     </tr>
     *     <tr>
     *         <td>8</td><td>10-8=2</td><numbers>5 3</numbers>
     *     </tr>
     *     <tr>
     *         <td>5</td><td>10-5=5</td><numbers>3</numbers>
     *     </tr>
     *
     * </table>
     *
     */
    private void buildSubProblems(){

        int numbersSize=numbers.size();

        /*
         * Numbers are supposed to be positive so if the target is negative,
         * there is no chance to find a valid solution.
         * As the list of numbers is sorted, the case when target < 0 happens quickly
         * Hence, it quickly removes combinations implying big numbers
         */
        if(target>=0 && numbersSize> 1){

            for(int i=0;i<numbersSize;i++){
                Integer nodeValue=numbers.get(i);
                List<Integer> subList=numbers.subList(i+1,numbersSize);
                int newTarget=this.target-nodeValue;
                Problem problem=new Problem(newTarget, subList, nodeValue, this);
                System.out.println("Created problem: "+problem.dump());
                this.listSubProblems.add(problem);
            }
        }
    }

    /**
     * @return True is the Problem contains exactly one number and that number equals the target.
     */
    public boolean isNodeSolution(){
        return this.target==0;
    }

    public Integer getNodeValue(){
        return this.nodeValue;
    }

    public List<Problem> getListSubProblems(){
        return this.listSubProblems;
    }

    public Problem getParentProblem(){
        return this.parentProblem;
    }

    public String dump(){
        StringBuilder sb=new StringBuilder();
        sb.append("{nodeValue: "+this.nodeValue);
        sb.append("; target: "+target);
        sb.append("; numbers:");
        for(Integer integer : numbers){
            sb.append(integer+",");
        }
        sb.append("}");
        sb.append("Valid? : "+ isNodeSolution());
        return sb.toString();
    }

}

下面是显示如何测试它的代码:

import java.util.Arrays;
import java.util.Collections;
import java.util.List;

public class Main {

    public static void main(String[] args) throws Exception{
        Integer numbers[]={1,3,8,2,7};
        int target=10;

        List<Integer> listNumbers= Arrays.asList(numbers);

        Collections.sort(listNumbers);
        Collections.reverse(listNumbers);

        //Build the root problem
        Problem problem=new Problem(target,listNumbers,null,null);

        //Solve it
        problem.solve();

        //Dump the result.
        dumpResult(problem);

        System.out.println("Done!");
    }

    private static void dumpResult(Problem problem){
        for(Problem p:problem.getListSubProblems()){
            if(p.isNodeSolution()){
                System.out.print("\nSolution :");
                dumpSolution(p);
            }
            dumpResult(p);
        }
    }

    private static void dumpSolution(Problem problem){
        //If the current node is not the root problem
        if(problem.getParentProblem()!=null){
            System.out.print(problem.getNodeValue() + ", ");
            dumpSolution(problem.getParentProblem());
        }
    }
}

这是一个输出示例:

Created problem: {nodeValue: 8; target: 2; numbers:7,3,2,1,}Valid? : false
Created problem: {nodeValue: 7; target: 3; numbers:3,2,1,}Valid? : false
Created problem: {nodeValue: 3; target: 7; numbers:2,1,}Valid? : false
Created problem: {nodeValue: 2; target: 8; numbers:1,}Valid? : false
Created problem: {nodeValue: 1; target: 9; numbers:}Valid? : false
Created problem: {nodeValue: 7; target: -5; numbers:3,2,1,}Valid? : false
Created problem: {nodeValue: 3; target: -1; numbers:2,1,}Valid? : false
Created problem: {nodeValue: 2; target: 0; numbers:1,}Valid? : true
Created problem: {nodeValue: 1; target: 1; numbers:}Valid? : false
Created problem: {nodeValue: 3; target: 0; numbers:2,1,}Valid? : true
Created problem: {nodeValue: 2; target: 1; numbers:1,}Valid? : false
Created problem: {nodeValue: 1; target: 2; numbers:}Valid? : false
Created problem: {nodeValue: 2; target: -2; numbers:1,}Valid? : false
Created problem: {nodeValue: 1; target: -1; numbers:}Valid? : false
Created problem: {nodeValue: 2; target: 5; numbers:1,}Valid? : false
Created problem: {nodeValue: 1; target: 6; numbers:}Valid? : false

Solution :2, 8,
Solution :3, 7, Done!

现在，这不包括隐含负数的初始问题。 要解决这种情况，请隔离所有负数并计算负数的所有组合，并求和。

然后，对于每个负数之和，创建一个仅包含正数和相应目标(初始目标 - 负数之和)的子问题

一种改进方法:问题的复杂程度取决于负数的组合数量。因此，如果负数多于正数，您只需反转所有值并解决反转问题。

另一种改进方法:您可以在内存中维护每个子问题的正数之和。如果 sum + nodeValue < target="">

关于java - 从集合中找出其和为零的子集？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/15532957/