理解硬币变化递归

Question

我试图用递归解决硬币兑换问题，发现了下面的代码

问题:给定无限种不同面值的硬币，计算硬币的数目。给定的数量可以由它们组成。

输入:

int[] coins = {1, 2}; 
int amount = 5;
int ways = change(amount, coins, coins.length - 1);
// expected ways = 3 --> 11111, 1112, 122

代码:

int change(int amount, int[] coins, int index) {
if (amount < 0) return 0;
if (amount == 0) return 1;

int ways = 0;
while (amount > 0 && index >= 0) {
ways += change(amount - coins[index], coins, index);
index = index - 1;
}
return ways;
}

我理解代码本身，也理解基本情况，但我无法理解它是如何封装递归/解决方案的。

。如果我在解factorial(n)，我可以说factorial(n) = n * factorial(n-1)，所以我可以清楚地看到;递归。在换硬币的例子中，我不能推导出类似的关系。有人能帮我一下吗?

Answer 1

递归行如下:ways += change(amount - coins[index], coins, index);

我已经注释了代码来解释一下。

//amount is the total value we want all our coins to add up to
//coins carries the values we can add up to get the amount
//index is the coin we're "on" right now - we'll explain this more in a bit
int change(int amount, int[] coins, int index) {
//we went too low: out last coin was too large and pushed us into negatives
if (amount < 0) return 0;
//exact change! we found a new way to make change with these coins
if (amount == 0) return 1;

//count the number of ways we can make the change
int ways = 0;
//here's where the recursion starts: we start at index, which is the number of
//coins available to us. in this case, we're going right to the end of the
//array to the "2" coin. we'll repeatedly subtract "2" from the amount until
//we hit 0, meaning we were able to meet the amount using only "2" coins, or
//we're unable to go any further.
//if we're unable to go further, we return one level up from the recursion,
//and decrease index by 1: this means we're now trying the "1" coin. 
//this process repeats, making as much change with the "2" coin as we can and
//falling back to the "1" coin when we get stuck or reach the bottom of the
//recursion.
while (amount > 0 && index >= 0) {
//try to use this same coin over and over, and when the function returns,
//whether through success or failure...
ways += change(amount - coins[index], coins, index);
//...move onto the next coin and repeat the process.
index = index - 1;
}

//the total number of times we were able to make exact change with these coins
return ways;
}

更简单地说:

desired value: 5    available coins: 1, 2
value = 5
coin = 2
5 - 2 = 3
. value = 3
. coin = 2
. 3 - 2 = 1
. . value = 1
. . coin = 2
. . 1 - 2 = -1 | fail
. . coin = 1
. . 1 - 1 = 0 | success
. . no more coins to try
. value = 3
. coin = 1
. 3 - 1 = 2
. . value = 2
. . coin = 1
. . 2 - 1 = 1
. . . value = 1
. . . coin = 1
. . . 1 - 1 = 0 | success
. . . no more coins to try
. . no more coins to try
. no more coins to try
value = 5
coin = 1
5 - 1 = 4
. value = 4
[...and so on]