[英]What is a good iterative solution for this recursive algorithm?
问题 :给定一个数组和一个目标编号,以数组中元素的唯一组合形式打印目标编号的书写方式 。
范例 :
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)
...
总和是对函数的每个递归调用的总和,其中减去数组值作为参数。
从概念上讲,递归可以表示为(讽刺地表示为迭代):
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,即使对于上面的琐碎情况,也会导致StackOverflowError。 如何最好地迭代以下解决方案:
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));
}
}
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];
这是Python中的解决方案。
#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)
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