使用动态规划进行硬币找零

我一直在使用动态规划来解决硬币找零问题。我尝试创建一个数组 fin[]，其中包含该索引所需的最小硬币数量，然后打印它。 我编写了一段代码，我认为应该给出正确的输出，但我不明白为什么它没有给出准确的答案。 例如：对于输入：4 3 1 2 3（4是要找的金额，3是可用硬币的种类数量，1 2 3是硬币价值列表）输出应该是： 0 1 1 1 2 （因为我们有 1,2,3 可用硬币，所以需要 0 个硬币来兑换 0，1 个硬币来兑换 1，1 个硬币来兑换 2，1 个硬币来兑换 3 和 2兑换硬币 4) 但它给出 0 1 2 2 2

这是代码：

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in= new Scanner(System.in);
        int ch = in.nextInt();
        int noc = in.nextInt();
        int[] ca = new int[noc];
        for(int i=0;i<noc;i++)
            {
                //taking input for coins available say a,b,c
            ca[i] = in.nextInt();
        }

       int[] fin = new int[ch+1]; //creating an array for 0 to change                                store the minimum number of coins required for each term at index

        int b=ch+1;
        for(int i=0;i<b;i++)
            {
            int count = i; //This initializes the min coins to that number so it is never greater than that number itself. (but I have a doubt here: what if we don't have 1 in coins available 

            for(int j=0; j<noc; j++)
                {
                int c = ca[j]; //this takes the value of coins available from starting everytime i changes

                if((c < i) && (fin[i-c] +1 < count)) // as we using dynamic programming it starts from base case, so the best value for each number i is stored in fin[] , when we check for number i+1, it checks best case for the previous numbers.
                    count = fin[i-c]+1 ;

            }
            fin[i]= count;
        }


        for(int i=0;i<b;i++)
            {
            System.out.println(fin[i]);
        }

    }
}

我从这个页面引用了：http://interactivepython.org/runestone/static/pythonds/Recursion/DynamicProgramming.html http://interactivepython.org/runestone/static/pythonds/Recursion/DynamicProgramming.html

有人可以帮忙吗？

The article http://interactivepython.org/runestone/static/pythonds/Recursion/DynamicProgramming.html您引用的，很好地解释了如何使用动态编程构建硬币兑换器算法。该算法的Python版本可以翻译成Java：

import java.util.Arrays;
import java.util.List;

public class CoinChanger {

    public int[] dpMakeChange(List<Integer> coinValueList, int change, int[] minCoins) {
        for (int cents = 0; cents <= change; cents++) {
            int coinCount = cents;
            for (Integer c : coinValueList) {
                if (c > cents) {
                    continue;
                }
                if (minCoins[cents - c] + 1 < coinCount) {
                    coinCount = minCoins[cents - c] + 1;
                }
            }
            minCoins[cents] = coinCount;
        }
        return minCoins;
    }

    public static void main(String[] args) {
        List<Integer> coinValueList = Arrays.asList(new Integer[]{1, 2, 3});
        int change = 10;
        int[] minCoins = new int[change + 1];
        int[] result = (new CoinChanger()).dpMakeChange(coinValueList, change, minCoins);
        for (int i = 0; i < result.length; i++) {
            System.out.println("For change = " + i + " number of coins = " + result[i]);
        }
    }
}

正如您的问题一样，使用价值 1、2 和 3 的硬币，之前的算法会为您提供正确的值：

For change = 0 number of coins = 0
For change = 1 number of coins = 1
For change = 2 number of coins = 1
For change = 3 number of coins = 1
For change = 4 number of coins = 2
For change = 5 number of coins = 2
For change = 6 number of coins = 2
For change = 7 number of coins = 3
For change = 8 number of coins = 3
For change = 9 number of coins = 3
For change = 10 number of coins = 4