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。



　　[b]Java[/b]

package com.algorithm.dynamicprogramming;

import java.util.Arrays;

/**
* 找零问题
* Created by yulinfeng on 7/5/17.
*/
public class Money {
public static void main(String[] args) {
int[] money = {1, 3, 2, 5};
int sum = 8;
int count = money(money, sum);
System.out.println(count);
}

private static int money(int[] money, int sum) {
int[] count = new int[sum + 1];
count[0] = 0;
for (int j = 1; j < sum + 1; j++) { //总金额数，1,2,3，……，sum
int minCoins = j;
for (int i = 0; i < money.length; i++) {    //遍历硬币的面值
if (j - money[i] >= 0) {
int temp = count[j - money[i]] + 1; //当前所需硬币数
if (temp < minCoins) {
minCoins = temp;
}
}
}

count[j] = minCoins;
}
System.out.println(Arrays.toString(count));
return count[sum];
}
}

　　Python3

#coding=utf-8
def charge_making(money, num):
'''
找零问题
'''
count = [0] * (num + 1)
count[0] = 0
for j in range(1, num + 1):
minCoins = j
for i in range(len(money)):
if j - money[i] >= 0:
temp = count[j - money[i]] + 1
if temp < minCoins:
minCoins = temp

count[j] = minCoins

return count[num]

money = [1, 3, 2, 5]
num = 8
count = charge_making(money, num)
print(count)

tag