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

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)

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