6

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

{(1,2,-3)}

{(1,-1)}

{(3,-3)}

{(5,-5)}

等等等等

我已经尝试过这段代码，但是当我设置target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);
    }
}
4

3 回答 3

2

这是一个解决方案的提议。

我首先解决了第一个子问题：我假设所有的数字和目标都是正数，然后我解决了真正的问题。

为此，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：排序列表：数字：8,7,3,2,1 目标：10 然后递归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括这个数的解（因为总和很快超过了目标）。

这是解决此子问题的注释代码：

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();
    }

}

这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，这不包括暗示负数的初始问题。要解决这种情况，请隔离所有负数并计算所有负数组合，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的子问题（初始目标 - 负数之和）

一种改进方法：问题的复杂性取决于负数组合的数量。因此，如果负数多于正数，您可以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 那么继续探索分支是没有用的。

于 2013-03-21T08:10:24.120 回答
1

我在大学期间在谷歌面试时遇到了这个问题，并在很长一段时间内解决了这个问题。

想一想，对于一个为 0 的集合，“必须”是负数，“必须有一组正数”。

脚步：

1. Created a 2 arrays negativeNumArrays and POsitiveNumArrays
2. Create a new negative set(does not allows duplicate) which is possible sums of     negative arrays ex -
    [-1,-2,-3] = [-1,-2,-3, {-1-2=3},{-1,-3=-4},{-2,-3=-5},{-6}] = [-1,-2,-3,-4,-5,-6]
So the set looked like
Key:Value
"1" =-1
"2" = -2
...
"2:3"=-5 
"1:2:3"=-6

Here 
"N6" = -6

3. For this new set of negative array find combination in positive 
   array which matches any of the 6 negative arrays.

Same as above say positive numbers are 3 and 4
So the set would look like
"3"=3

"4"=4

"3:4"=7


Now simple compare the two sets and see which of these are equal
So for example Negative Set "1:3" = Positive Set "4"
and hence use Stringtokenizer to get the numbers from set key {-1,-3,4}
于 2013-03-20T20:12:25.157 回答
0

您没有检查是否partial为空，在这种情况下，当== 0sum_up_recursive()时将立即返回第一次尝试。试试这个：target

if (partial.size() > 0) {
    for (int x : partial)
        s += x;

    if (s == target)
        System.out.println("sum(" + Arrays.toString(partial.toArray()) + ")=" + target);

    if (s >= target)
        return;
}

请注意，可能还有其他方法可以极大地改进您正在使用的算法。我只是在回答为什么您的代码没有按预期工作。

于 2013-03-20T22:24:52.133 回答