你如何从给定的N个数字中测试所有可能的加法组合,使它们加起来得到给定的最终数字?

一个简单的例子:

要添加的数字集:N ={1,5,22,15,0,…} 期望结果:12345


当前回答

Thank you.. ephemient

我已经将上述逻辑从python转换为php..

<?php
$data = array(array(2,3,5,10,15),array(4,6,23,15,12),array(23,34,12,1,5));
$maxsum = 25;

print_r(bestsum($data,$maxsum));  //function call

function bestsum($data,$maxsum)
{
$res = array_fill(0, $maxsum + 1, '0');
$res[0] = array();              //base case
foreach($data as $group)
{
 $new_res = $res;               //copy res

  foreach($group as $ele)
  {
    for($i=0;$i<($maxsum-$ele+1);$i++)
    {   
        if($res[$i] != 0)
        {
            $ele_index = $i+$ele;
            $new_res[$ele_index] = $res[$i];
            $new_res[$ele_index][] = $ele;
        }
    }
  }

  $res = $new_res;
}

 for($i=$maxsum;$i>0;$i--)
  {
    if($res[$i]!=0)
    {
        return $res[$i];
        break;
    }
  }
return array();
}
?>

其他回答

另一个python解决方案是使用itertools.combination模块,如下所示:

#!/usr/local/bin/python

from itertools import combinations

def find_sum_in_list(numbers, target):
    results = []
    for x in range(len(numbers)):
        results.extend(
            [   
                combo for combo in combinations(numbers ,x)  
                    if sum(combo) == target
            ]   
        )   

    print results

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

输出:[(8,7),(5,10),(3,8,4),(3,5,7)]

PHP版本,灵感来自Keith Beller的c#版本。

bala的PHP版本不适合我,因为我不需要对数字进行分组。我想要一个更简单的实现,只有一个目标值和一个数字池。这个函数也会删除任何重复的条目。

编辑25/10/2021:添加精度参数以支持浮点数(现在需要bcmath扩展)。

/**
 * Calculates a subset sum: finds out which combinations of numbers
 * from the numbers array can be added together to come to the target
 * number.
 *
 * Returns an indexed array with arrays of number combinations.
 *
 * Example:
 *
 * <pre>
 * $matches = subset_sum(array(5,10,7,3,20), 25);
 * </pre>
 *
 * Returns:
 *
 * <pre>
 * Array
 * (
 *   [0] => Array
 *   (
 *       [0] => 3
 *       [1] => 5
 *       [2] => 7
 *       [3] => 10
 *   )
 *   [1] => Array
 *   (
 *       [0] => 5
 *       [1] => 20
 *   )
 * )
 * </pre>
 *
 * @param number[] $numbers
 * @param number $target
 * @param array $part
 * @param int $precision
 * @return array[number[]]
 */
function subset_sum($numbers, $target, $precision=0, $part=null)
{
    // we assume that an empty $part variable means this
    // is the top level call.
    $toplevel = false;
    if($part === null) {
        $toplevel = true;
        $part = array();
    }

    $s = 0;
    foreach($part as $x)
    {
        $s = $s + $x;
    }

    // we have found a match!
    if(bccomp((string) $s, (string) $target, $precision) === 0)
    {
        sort($part); // ensure the numbers are always sorted
        return array(implode('|', $part));
    }

    // gone too far, break off
    if($s >= $target)
    {
        return null;
    }

    $matches = array();
    $totalNumbers = count($numbers);

    for($i=0; $i < $totalNumbers; $i++)
    {
        $remaining = array();
        $n = $numbers[$i];

        for($j = $i+1; $j < $totalNumbers; $j++)
        {
            $remaining[] = $numbers[$j];
        }

        $part_rec = $part;
        $part_rec[] = $n;

        $result = subset_sum($remaining, $target, $precision, $part_rec);
        if($result)
        {
            $matches = array_merge($matches, $result);
        }
    }

    if(!$toplevel)
    {
        return $matches;
    }

    // this is the top level function call: we have to
    // prepare the final result value by stripping any
    // duplicate results.
    $matches = array_unique($matches);
    $result = array();
    foreach($matches as $entry)
    {
        $result[] = explode('|', $entry);
    }

    return $result;
}

例子:

$result = subset_sum(array(5, 10, 7, 3, 20), 25);

这将返回一个包含两个数字组合数组的索引数组:

3, 5, 7, 10
5, 20

浮点数示例:

// Specify the precision in the third argument
$result = subset_sum(array(0.40, 0.03, 0.05), 0.45, 2);

这将返回一个匹配项:

0.40, 0.05

c#版本的@msalvadores代码的答案

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

static void sum_up_recursive(List<int> numbers, int target, List<int> part)
{
   int s = 0;
   foreach (int x in part)
   {
       s += x;
   }
   if (s == target)
   {
        Console.WriteLine("sum(" + string.Join(",", part.Select(n => n.ToString()).ToArray()) + ")=" + target);
   }
   if (s >= target)
   {
        return;
   }
   for (int i = 0;i < numbers.Count;i++)
   {
         var remaining = new List<int>();
         int n = numbers[i];
         for (int j = i + 1; j < numbers.Count;j++)
         {
             remaining.Add(numbers[j]);
         }
         var part_rec = new List<int>(part);
         part_rec.Add(n);
         sum_up_recursive(remaining,target,part_rec);
   }
}
static void sum_up(List<int> numbers, int target)
{
    sum_up_recursive(numbers,target,new List<int>());
}

用excel找到组合(相当容易)。 (你的电脑不能太慢)

去这个网站 进入“Sum to Target”页面 下载“Sum to Target”excel文件。 按照网站页面上的说明操作。

希望这能有所帮助。

func sum(array : [Int]) -> Int{
    var sum = 0
    array.forEach { (item) in
        sum = item + sum
    }
    return sum
}
func susetNumbers(array :[Int], target : Int, subsetArray: [Int],result : inout [[Int]]) -> [[Int]]{
    let s = sum(array: subsetArray)
    if(s == target){
        print("sum\(subsetArray) = \(target)")
        result.append(subsetArray)
    }
    for i in 0..<array.count{
        let n = array[i]
        let remaning = Array(array[(i+1)..<array.count])
        susetNumbers(array: remaning, target: target, subsetArray: subsetArray + [n], result: &result)
        
    }
    return result
}

 var resultArray = [[Int]]()
    let newA = susetNumbers(array: [1,2,3,4,5], target: 5, subsetArray: [],result:&resultArray)
    print(resultArray)