Algorithm 这是正确的重复关系I'；你为硬币兑换挑战找到了什么？

我试图解决“硬币兑换问题”，我想我已经想出了一个递归的解决方案，但我想验证一下

作为一个例子，让我们假设我们有便士、五分镍币和一角镍币，并试图用22美分兑换

C = { 1 = penny, nickle = 5, dime = 10 }
K = 22 

那么做出改变的方法有很多

f(C,N) = f({1,5,10},22) 
=
  (# of ways to make change with 0 dimes) 
+ (# of ways to make change with 1 dimes) 
+ (# of ways to make change with 2 dimes) 

= f(C\{dime},22-0*10) + f(C\{dime},22-1*10) + f(C\{dime},22-2*10)
= f({1,5},22) + f({1,5},12) + f({1,5},2)

及

等等

换句话说，我的算法是

let f(C,K) be the number of ways to make change for K cents with coins C
and have the following implementation

if(C is empty or K=0)
    return 0
sum = 0
m = C.PopLargest()
A = {0, 1, ..., K / m}
for(i in A)
   sum += f(C,K-i*m)
return sum

如果有任何缺陷呢

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我想是线性时间

重新思考您的基本案例：

1. What if K < 0 ? Then no solution exists. i.e. No of ways = 0.

2. When K = 0, so there is 1 way to make changes and which is to consider zero elements from array of coin-types. 

3. When coin array is empty then No of ways = 0.

这将是N（数组大小）的指数形式

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如果您希望改进，我建议您使用动态规划。

我的回答对您有帮助吗？

1. What if K < 0 ? Then no solution exists. i.e. No of ways = 0.

2. When K = 0, so there is 1 way to make changes and which is to consider zero elements from array of coin-types. 

3. When coin array is empty then No of ways = 0.

T(N) = O(N) + K*T(N-1)