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找出所有可能的数字组合以达到给定的总和

发布于 2024-10-10 19:37:07 字数 175 浏览 10 评论 0原文

您将如何测试给定N 组数字中所有可能的加法组合，以便它们相加达到给定的最终数字？

一个简短的示例：

要添加的一组数字：N = {1,5,22,15,0,...}
所需结果：12345

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你的呼吸 2024-10-17 19:37:07

这个问题可以通过所有可能的总和的递归组合过滤掉达到目标的总和来解决。下面是 Python 中的算法：

def subset_sum(numbers, target, partial=[]):
    s = sum(partial)

    # check if the partial sum is equals to target
    if s == target: 
        print "sum(%s)=%s" % (partial, target)
    if s >= target:
        return  # if we reach the number why bother to continue
    
    for i in range(len(numbers)):
        n = numbers[i]
        remaining = numbers[i+1:]
        subset_sum(remaining, target, partial + [n]) 
   

if __name__ == "__main__":
    subset_sum([3,9,8,4,5,7,10],15)

    #Outputs:
    #sum([3, 8, 4])=15
    #sum([3, 5, 7])=15
    #sum([8, 7])=15
    #sum([5, 10])=15

这种类型的算法在下面的斯坦福抽象编程讲座<中得到了很好的解释。 /a> - 非常推荐此视频来了解递归如何生成解决方案的排列。

编辑

以上作为生成器函数，使其更有用一点。由于yield from，需要 Python 3.3+。

def subset_sum(numbers, target, partial=[], partial_sum=0):
    if partial_sum == target:
        yield partial
    if partial_sum >= target:
        return
    for i, n in enumerate(numbers):
        remaining = numbers[i + 1:]
        yield from subset_sum(remaining, target, partial + [n], partial_sum + n)

这是相同算法的 Java 版本：

package tmp;

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 = {3,9,8,4,5,7,10};
        int target = 15;
        sum_up(new ArrayList<Integer>(Arrays.asList(numbers)),target);
    }
}

它是完全相同的启发式算法。我的Java有点生疏，但我认为很容易理解。

Java 解决方案的 C# 转换： （作者：@JeremyThompson）

public static void Main(string[] args)
{
    List<int> numbers = new List<int>() { 3, 9, 8, 4, 5, 7, 10 };
    int target = 15;
    sum_up(numbers, target);
}

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

private static void sum_up_recursive(List<int> numbers, int target, List<int> partial)
{
    int s = 0;
    foreach (int x in partial) s += x;

    if (s == target)
        Console.WriteLine("sum(" + string.Join(",", partial.ToArray()) + ")=" + target);

    if (s >= target)
        return;

    for (int i = 0; i < numbers.Count; i++)
    {
        List<int> remaining = new List<int>();
        int n = numbers[i];
        for (int j = i + 1; j < numbers.Count; j++) remaining.Add(numbers[j]);

        List<int> partial_rec = new List<int>(partial);
        partial_rec.Add(n);
        sum_up_recursive(remaining, target, partial_rec);
    }
}

Ruby 解决方案： （作者：@emaillenin）

def subset_sum(numbers, target, partial=[])
  s = partial.inject 0, :+
# check if the partial sum is equals to target

  puts "sum(#{partial})=#{target}" if s == target

  return if s >= target # if we reach the number why bother to continue

  (0..(numbers.length - 1)).each do |i|
    n = numbers[i]
    remaining = numbers.drop(i+1)
    subset_sum(remaining, target, partial + [n])
  end
end

subset_sum([3,9,8,4,5,7,10],15)

编辑：复杂性讨论

正如其他人提到的，这是一个 NP 难题。它可以在指数时间 O(2^n) 内解决，例如，对于 n=10，将有 1024 个可能的解决方案。如果您尝试达到的目标处于较低范围内，则此算法有效。例如：

subset_sum([1,2,3,4,5,6,7,8,9,10],100000) 生成 1024 个分支，因为目标永远无法过滤掉可能的解决方案。

另一方面 subset_sum([1,2,3,4,5,6,7,8,9,10],10) 仅生成 175 个分支，因为要达到 的目标10 可以过滤掉许多组合。

如果 N 和 Target 是大数，则应该采用解决方案的近似版本。

This problem can be solved with a recursive combinations of all possible sums filtering out those that reach the target. Here is the algorithm in Python:

def subset_sum(numbers, target, partial=[]):
    s = sum(partial)

    # check if the partial sum is equals to target
    if s == target: 
        print "sum(%s)=%s" % (partial, target)
    if s >= target:
        return  # if we reach the number why bother to continue
    
    for i in range(len(numbers)):
        n = numbers[i]
        remaining = numbers[i+1:]
        subset_sum(remaining, target, partial + [n]) 
   

if __name__ == "__main__":
    subset_sum([3,9,8,4,5,7,10],15)

    #Outputs:
    #sum([3, 8, 4])=15
    #sum([3, 5, 7])=15
    #sum([8, 7])=15
    #sum([5, 10])=15

This type of algorithms are very well explained in the following Stanford's Abstract Programming lecture - this video is very recommendable to understand how recursion works to generate permutations of solutions.

Edit

The above as a generator function, making it a bit more useful. Requires Python 3.3+ because of yield from.

def subset_sum(numbers, target, partial=[], partial_sum=0):
    if partial_sum == target:
        yield partial
    if partial_sum >= target:
        return
    for i, n in enumerate(numbers):
        remaining = numbers[i + 1:]
        yield from subset_sum(remaining, target, partial + [n], partial_sum + n)

Here is the Java version of the same algorithm:

package tmp;

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 = {3,9,8,4,5,7,10};
        int target = 15;
        sum_up(new ArrayList<Integer>(Arrays.asList(numbers)),target);
    }
}

It is exactly the same heuristic. My Java is a bit rusty but I think is easy to understand.

C# conversion of Java solution: (by @JeremyThompson)

public static void Main(string[] args)
{
    List<int> numbers = new List<int>() { 3, 9, 8, 4, 5, 7, 10 };
    int target = 15;
    sum_up(numbers, target);
}

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

private static void sum_up_recursive(List<int> numbers, int target, List<int> partial)
{
    int s = 0;
    foreach (int x in partial) s += x;

    if (s == target)
        Console.WriteLine("sum(" + string.Join(",", partial.ToArray()) + ")=" + target);

    if (s >= target)
        return;

    for (int i = 0; i < numbers.Count; i++)
    {
        List<int> remaining = new List<int>();
        int n = numbers[i];
        for (int j = i + 1; j < numbers.Count; j++) remaining.Add(numbers[j]);

        List<int> partial_rec = new List<int>(partial);
        partial_rec.Add(n);
        sum_up_recursive(remaining, target, partial_rec);
    }
}

Ruby solution: (by @emaillenin)

def subset_sum(numbers, target, partial=[])
  s = partial.inject 0, :+
# check if the partial sum is equals to target

  puts "sum(#{partial})=#{target}" if s == target

  return if s >= target # if we reach the number why bother to continue

  (0..(numbers.length - 1)).each do |i|
    n = numbers[i]
    remaining = numbers.drop(i+1)
    subset_sum(remaining, target, partial + [n])
  end
end

subset_sum([3,9,8,4,5,7,10],15)

Edit: complexity discussion

As others mention this is an NP-hard problem. It can be solved in exponential time O(2^n), for instance for n=10 there will be 1024 possible solutions. If the targets you are trying to reach are in a low range then this algorithm works. So for instance:

subset_sum([1,2,3,4,5,6,7,8,9,10],100000) generates 1024 branches because the target never gets to filter out possible solutions.

On the other hand subset_sum([1,2,3,4,5,6,7,8,9,10],10) generates only 175 branches, because the target to reach 10 gets to filter out many combinations.

If N and Target are big numbers one should move into an approximate version of the solution.

找出所有可能的数字组合以达到给定的总和

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