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，……，sum21 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 = j10 for i in range(len(money)):11 if j - money[i] >= 0:12 temp = count[j - money[i]] + 113 if temp < minCoins:14 minCoins = temp15 16 count[j] = minCoins17 18 return count[num]19 20 money = [1, 3, 2, 5]21 num = 822 count = charge_making(money, num)23 print(count)
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