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我一直在使用动态编程解决硬币找零问题。我试图创建一个数组 fin[] ，其中包含该索引所需的最小硬币数量，然后打印它。我写了一个我认为应该给出正确输出的代码，但我不知道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 是改变的数量，3 是可用硬币类型的数量，1 2 3 是硬币值的列表） 输出应该是：0 1 1 1 2（因为我们有 1,2,3 作为可用硬币，它需要 0 个硬币来改变 0，1 个硬币来改变 1，1 个硬币来改变 2，1 个硬币来改变 3 和 2 个硬币来改变 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]);
        }

    }
}

我参考了这个页面：http: //interactivepython.org/runestone/static/pythonds/Recursion/DynamicProgramming.html

任何人都可以帮忙吗？

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1 回答 1

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您引用的文章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版本可以用Java翻译：

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]);
        }
    }
}

如您的问题所示，使用价值 1、2 和 3 的硬币，前面的算法为您提供了正确的值：

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
于 2015-10-07T15:39:49.263 回答