9.8 Given an infinite number of quarters (25 cents), dimes (10 cents), nickels (5 cents) and pennies (1 cent), write code to calculate the number of ways of representing n cents.

这道题给定一个钱数,让我们求用quarter,dime,nickle和penny来表示的方法总和,很明显还是要用递归来做。比如我们有50美分,那么

makeChange(50) =

  makeChange(50 using 0 quarter) +

  makeChange(50 using 1 quarter) +

  makeChange(50 using 2 quarters)

而其中第一个makeChange(50 using 0 quarter)又可以拆分为:

makeChange(50 using 0 quarter) =

  makeChange(50 using 0 quarter, 0 dimes) +

  makeChange(50 using 0 quarter, 1 dimes) +

  makeChange(50 using 0 quarter, 2 dimes) +

  makeChange(50 using 0 quarter, 3 dimes) +

  makeChange(50 using 0 quarter, 4 dimes) +

  makeChange(50 using 0 quarter, 5 dimes)

而这里面的每项又可以继续往下拆成nickle和penny,整体是一个树形结构,计算顺序是从最底层开始,也就是给定的钱数都是由penny组成的情况慢慢往回递归,加一个nickle,加两个nickle,再到加dime和quarter,参见代码如下:

解法一:.

class Solution {
public:
int makeChange(int n) {
vector<int> denoms = {, , , };
return makeChange(n, denoms, );
}
int makeChange(int amount, vector<int> denoms, int idx) {
if (idx >= denoms.size() - ) return ;
int val = denoms[idx], res = ;
for (int i = ; i * val <= amount; ++i) {
int rem = amount - i * val;
res += makeChange(rem, denoms, idx + );
}
return res;
}
};

上述代码虽然正确但是效率一般,因为存在大量的重复计算,我们可以用哈希表来保存计算过程中的结果,下次遇到相同结果时,直接从哈希表中取出来即可,参见代码如下:

解法二:

class Solution {
public:
int makeChange(int n) {
vector<int> denoms = {, , , };
vector<vector<int> > m(n + , vector<int>(denoms.size()));
return makeChange(n, denoms, , m);
}
int makeChange(int amount, vector<int> denoms, int idx, vector<vector<int> > &m) {
if (m[amount][idx] > ) return m[amount][idx];
if (idx >= denoms.size() - ) return ;
int val = denoms[idx], res = ;
for (int i = ; i * val <= amount; ++i) {
int rem = amount - i * val;
res += makeChange(rem, denoms, idx + , m);
}
m[amount][idx] = res;
return res;
}
};

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