对于这种递归算法，什么是好的迭代解决方案？

Question

Problem：给定数组和目标号，打印目标数可以作为数组中元素的唯一组合的方式。

示例：

array = {1,2,3} target = 4
4 = {2,2}, {3,1}, {1,1,1,1} //numbers can be repeatedly selected from the array.
Ans: 3 ways

递归解

F(4) = F(4-1) + F(4-2) + F(4-3)
F(4) = F(3) + F(2) + F(1)
...

总计和是对函数的每个递归调用的和，其中数组值减去作为参数。

概念上的重复可以表示为(具有讽刺意味的是，作为迭代)：

for(int i=0; i<array.length; i++)
   sum+=F(target - array[i]);

基本案例：

F(0) = 1 //Implies sums to target
F(<0) = 0 //Implies cannot sum to target

但是，即使对于上面这样的简单情况，它也会导致一个StackOverflowError。如何最好地迭代下面的解决方案：

码

public class CombinationSum {

    private int[] array;
    private int target;

    public CombinationSum(int[] array, int target)
    {
        this.array = array;
        this.target = target;
    }

    public int recurSum(int val)
    {
        int sum = 0;

        if(val < 0 )
            return 0;
        else if(val == 0)
            return 1;
        else 
        {
            for(int i = 0; i<array.length; i++)
            {
                sum+= recurSum(target-array[i]); //heavy recursion here?

            }

            return sum;
        }
    }

    public static void main(String[] args)
    {
        int target = 4;
        int[] array = {1,2,3};

        CombinationSum cs = new CombinationSum(array, target);

        System.out.println("Number of possible combinations: "+cs.recurSum(target));
    }

}

Answer 1

伪码

F[0]=1;

for(i=1;i<=n;++i){
  F[i]=0;
  for(j=1;j<i++j){
    F[i]+=F[i-j];
  }

}

print F[n];

Answer 2

下面是Python中的一个解决方案。

#input is a list A of positive integers, and a target integer n
#output is a list of sublists of A (repeats OK!) with sum n
#permutations will appear
def make_list(A,n):
    A = list(set(A)) #eliminate repeats in A
    L = []
    if n > 0:
        for a in A:
            if type(make_list(A,n-a)) == list:
                for l in make_list(A,n-a):
                    l.append(a)
                    L.append(l)
    elif n == 0:
        L = [[]]
    else: L = -1
    return L

#returns the count of distinct lists in make_list(A,n)
def counter(A,n):
    b = []
    for l in make_list(A,n):    
        if sorted(l) not in b:
            b.append(sorted(l))
    return len(b)