11.动态规划（4）——找零问题

　　找零问题：需找零金额为W，硬币面值有(d1, d2, d3，…，dm)，最少需要多少枚硬币。

　　问题：需找零金额为8，硬币面值有(1, 3, 2, 5)，最少需要多少枚硬币。

　　设F(j)表示总金额为j时最少的零钱数，F(0) = 0，W表示找零金额，有零钱一堆{d1, d2, d3，…，dm}。同样根据之前的经验，要达到为j，那么必然是j – di(1 <= i <= m)面值的硬币数再加1个di面值的硬币，当然j >= di，即F(j) = F(j - di) + 1, j >= di。

　　Java

 1 package com.algorithm.dynamicprogramming;
 2
 3 import java.util.Arrays;
 4
 5 /**
 6  * 找零问题
 7  * Created by yulinfeng on 7/5/17.
 8  */
 9 public class Money {
10     public static void main(String[] args) {
11         int[] money = {1, 3, 2, 5};
12         int sum = 8;
13         int count = money(money, sum);
14         System.out.println(count);
15     }
16
17     private static int money(int[] money, int sum) {
18         int[] count = new int[sum + 1];
19         count[0] = 0;
20         for (int j = 1; j < sum + 1; j++) { //总金额数，1,2,3，……，sum
21             int minCoins = j;
22             for (int i = 0; i < money.length; i++) {    //遍历硬币的面值
23                 if (j - money[i] >= 0) {
24                     int temp = count[j - money[i]] + 1; //当前所需硬币数
25                     if (temp < minCoins) {
26                         minCoins = temp;
27                     }
28                 }
29             }
30
31             count[j] = minCoins;
32         }
33         System.out.println(Arrays.toString(count));
34         return count[sum];
35     }
36 }

　　Python3

 1 #coding=utf-8
 2 def charge_making(money, num):
 3     ‘‘‘
 4     找零问题
 5     ‘‘‘
 6     count = [0] * (num + 1)
 7     count[0] = 0
 8     for j in range(1, num + 1):
 9         minCoins = j
10         for i in range(len(money)):
11             if j - money[i] >= 0:
12                 temp = count[j - money[i]] + 1
13                 if temp < minCoins:
14                     minCoins = temp
15
16         count[j] = minCoins
17
18     return count[num]
19
20 money = [1, 3, 2, 5]
21 num = 8
22 count = charge_making(money, num)
23 print(count)

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