[CareerCup] 9.8 Represent N Cents 美分的组成
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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