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

我将c#示例移植到Objective-c,并没有在响应中看到它:

//Usage
NSMutableArray* numberList = [[NSMutableArray alloc] init];
NSMutableArray* partial = [[NSMutableArray alloc] init];
int target = 16;
for( int i = 1; i<target; i++ )
{ [numberList addObject:@(i)]; }
[self findSums:numberList target:target part:partial];


//*******************************************************************
// Finds combinations of numbers that add up to target recursively
//*******************************************************************
-(void)findSums:(NSMutableArray*)numbers target:(int)target part:(NSMutableArray*)partial
{
    int s = 0;
    for (NSNumber* x in partial)
    { s += [x intValue]; }

    if (s == target)
    { NSLog(@"Sum[%@]", partial); }

    if (s >= target)
    { return; }

    for (int i = 0;i < [numbers count];i++ )
    {
        int n = [numbers[i] intValue];
        NSMutableArray* remaining = [[NSMutableArray alloc] init];
        for (int j = i + 1; j < [numbers count];j++)
        { [remaining addObject:@([numbers[j] intValue])]; }

        NSMutableArray* partRec = [[NSMutableArray alloc] initWithArray:partial];
        [partRec addObject:@(n)];
        [self findSums:remaining target:target part:partRec];
    }
}

其他回答

我在做类似的scala作业。我想在这里发布我的解决方案:

 def countChange(money: Int, coins: List[Int]): Int = {
      def getCount(money: Int, remainingCoins: List[Int]): Int = {
        if(money == 0 ) 1
        else if(money < 0 || remainingCoins.isEmpty) 0
        else
          getCount(money, remainingCoins.tail) +
            getCount(money - remainingCoins.head, remainingCoins)
      }
      if(money == 0 || coins.isEmpty) 0
      else getCount(money, coins)
    }

Java解决方案的Swift 3转换(by @JeremyThompson)

protocol _IntType { }
extension Int: _IntType {}


extension Array where Element: _IntType {

    func subsets(to: Int) -> [[Element]]? {

        func sum_up_recursive(_ numbers: [Element], _ target: Int, _ partial: [Element], _ solution: inout [[Element]]) {

            var sum: Int = 0
            for x in partial {
                sum += x as! Int
            }

            if sum == target {
                solution.append(partial)
            }

            guard sum < target else {
                return
            }

            for i in stride(from: 0, to: numbers.count, by: 1) {

                var remaining = [Element]()

                for j in stride(from: i + 1, to: numbers.count, by: 1) {
                    remaining.append(numbers[j])
                }

                var partial_rec = [Element](partial)
                partial_rec.append(numbers[i])

                sum_up_recursive(remaining, target, partial_rec, &solution)
            }
        }

        var solutions = [[Element]]()
        sum_up_recursive(self, to, [Element](), &solutions)

        return solutions.count > 0 ? solutions : nil
    }

}

用法:

let numbers = [3, 9, 8, 4, 5, 7, 10]

if let solution = numbers.subsets(to: 15) {
    print(solution) // output: [[3, 8, 4], [3, 5, 7], [8, 7], [5, 10]]
} else {
    print("not possible")
}

我将c#示例移植到Objective-c,并没有在响应中看到它:

//Usage
NSMutableArray* numberList = [[NSMutableArray alloc] init];
NSMutableArray* partial = [[NSMutableArray alloc] init];
int target = 16;
for( int i = 1; i<target; i++ )
{ [numberList addObject:@(i)]; }
[self findSums:numberList target:target part:partial];


//*******************************************************************
// Finds combinations of numbers that add up to target recursively
//*******************************************************************
-(void)findSums:(NSMutableArray*)numbers target:(int)target part:(NSMutableArray*)partial
{
    int s = 0;
    for (NSNumber* x in partial)
    { s += [x intValue]; }

    if (s == target)
    { NSLog(@"Sum[%@]", partial); }

    if (s >= target)
    { return; }

    for (int i = 0;i < [numbers count];i++ )
    {
        int n = [numbers[i] intValue];
        NSMutableArray* remaining = [[NSMutableArray alloc] init];
        for (int j = i + 1; j < [numbers count];j++)
        { [remaining addObject:@([numbers[j] intValue])]; }

        NSMutableArray* partRec = [[NSMutableArray alloc] initWithArray:partial];
        [partRec addObject:@(n)];
        [self findSums:remaining target:target part:partRec];
    }
}

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>());
}

下面是一个Java版本,它非常适合小N和非常大的目标和,当复杂度O(t*N)(动态解)大于指数算法时。我的版本在中间攻击中使用了一个meet,并进行了一些调整,以降低复杂度,从经典的naive O(n*2^n)降低到O(2^(n/2))。

如果你想在32到64个元素之间的集合中使用这种方法,你应该将表示step函数中当前子集的int改为long,尽管随着集合大小的增加,性能显然会急剧下降。如果你想对一个有奇数个元素的集合使用这个,你应该给这个集合加上一个0,使它成为偶数。

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

public class SubsetSumMiddleAttack {
    static final int target = 100000000;
    static final int[] set = new int[]{ ... };

    static List<Subset> evens = new ArrayList<>();
    static List<Subset> odds = new ArrayList<>();

    static int[][] split(int[] superSet) {
        int[][] ret = new int[2][superSet.length / 2]; 

        for (int i = 0; i < superSet.length; i++) ret[i % 2][i / 2] = superSet[i];

        return ret;
    }

    static void step(int[] superSet, List<Subset> accumulator, int subset, int sum, int counter) {
        accumulator.add(new Subset(subset, sum));
        if (counter != superSet.length) {
            step(superSet, accumulator, subset + (1 << counter), sum + superSet[counter], counter + 1);
            step(superSet, accumulator, subset, sum, counter + 1);
        }
    }

    static void printSubset(Subset e, Subset o) {
        String ret = "";
        for (int i = 0; i < 32; i++) {
            if (i % 2 == 0) {
                if ((1 & (e.subset >> (i / 2))) == 1) ret += " + " + set[i];
            }
            else {
                if ((1 & (o.subset >> (i / 2))) == 1) ret += " + " + set[i];
            }
        }
        if (ret.startsWith(" ")) ret = ret.substring(3) + " = " + (e.sum + o.sum);
        System.out.println(ret);
    }

    public static void main(String[] args) {
        int[][] superSets = split(set);

        step(superSets[0], evens, 0,0,0);
        step(superSets[1], odds, 0,0,0);

        for (Subset e : evens) {
            for (Subset o : odds) {
                if (e.sum + o.sum == target) printSubset(e, o);
            }
        }
    }
}

class Subset {
    int subset;
    int sum;

    Subset(int subset, int sum) {
        this.subset = subset;
        this.sum = sum;
    }
}