使用动态规划的硬币兑换

Question

我一直在使用动态规划研究硬币兑换问题。我尝试制作一个数组fin[]，其中包含索引所需的最小硬币数，然后打印它。我已经写了一个代码，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是要更改的金额，3是可用硬币的类型，1 2 3是硬币值列表）输出应该是：0 1 1 2（因为我们有1,2,3作为可用硬币，需要0个硬币更改0，1个硬币更改1，1个硬币更改2，1个硬币更改3和2个硬币更改4）但它给出了0 1 2

代码如下：

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]);
        }

    }
}

我参考了这一页：http://interactivepython.org/runestone/static/pythonds/Recursion/DynamicProgramming.html

有人能帮忙吗？

Answer 1

你引用的那篇文章很好地解释了如何使用动态规划构建硬币兑换器算法。该算法的python版本可以翻译为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]);
        }
    }
}

使用价值1、2和3的硬币，就像你的问题一样，前面的算法会给你正确的值：

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