TDD（练习）Count Coins 兑换一美元的方法

问题描述

There are four types of common coins in US currency:
quarters (25 cents)
dimes (10 cents)
nickels (5 cents)
pennies (1 cent)

There are 6 ways to make change for 15 cents:
A dime and a nickel;
A dime and 5 pennies;
3 nickels;
2 nickels and 5 pennies;
A nickel and 10 pennies;
15 pennies.

How many ways are there to make change for a dollar
using these common coins? (1 dollar = 100 cents).

问题说明

1 dollar = 100 cents ，现在有一美元需要兑换零钱。上面说明有四种单位， quarters (25 cents) dimes (10 cents) nickels (5 cents) pennies (1 cent)。 可以参看上面15 cents 有6种兑换方案。需要计算 1dollar 一共有多少种兑换方案 。
兑换方案举例：
100pennies 是一种
2 quarters + 50pennies 是一种

看到这道题后总是逃不过原来的习惯---想，一定要全部想明白了再做。结果导致想了半天，也没有开始做。看起Bob Deng 说的没错，需要每天一小时。TDD是需要刻意训练的。

测试用例

import static org.junit.Assert.assertEquals;

public class CountCoinsTest {
    int[] coins = {25,10,5,1};
    @Test
    public void should_0_0(){
        int count = CountCoins.makeChange(0);
        assertEquals(0,count);
    }

    @Test
    public void should_2_5(){
        int count = CountCoins.makeChange(5);
        assertEquals(2,count);
    }
    @Test
    public void should_6_15(){
        int count = CountCoins.makeChange(15);
        assertEquals(6,count);
    }
    @Test
    public void should_13_25(){
        int count = CountCoins.makeChange(25);
        assertEquals(13,count);
    }

    @Test
    public void should_18_30(){
        int count = CountCoins.makeChange(30);
        assertEquals(18,count);
    }
    @Test
    public void should_242_100(){
        int count = CountCoins.makeChange(100);
        assertEquals(242,count);
    }

   
}

实现类

public class CountCoins {

    public static int makeChange(int amount) {
        int ways = 0;
        if (amount < 1) return 0;
        for (int quarters = 0; quarters <= amount; quarters += 25) {
            for (int dimes = 0; dimes <= amount; dimes += 10) {
                for (int nickels = 0; nickels <= amount; nickels += 5) {

                    int pennies = amount - quarters - dimes - nickels;
                    if (pennies < 0) break;
//                    System.out.println(quarters + "  " + dimes + "  " + nickels + " " + pennies);
                    ways++;

//                    for (int pennies = 0; pennies <= amount; pennies++) {
//
//                        if (quarters + dimes + nickels + pennies == amount) {
//                            System.out.println(quarters + "  " + dimes + "  " + nickels + " " + pennies);
//                            ways++;
//                            break;
//                        }
//                    }
                }
            }
        }

        return ways;
    }
}

以上是对该题的一种简单的实现方法，比较直观。也可以参考武可写的，远程异步结对 - Count Coins 可以对本题以及TDD有更好的认识。