You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.
Note: You can assume that
- 0 <= amount <= 5000
- 1 <= coin <= 5000
- the number of coins is less than 500
- the answer is guaranteed to fit into signed 32-bit integer
Example 1:
Input: amount = 5, coins = [1, 2, 5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1
Example 2:
Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.
Example 3:
Input: amount = 10, coins = [10]
Output: 1
题解:
第一次WA:
想法是遍历coins,然后降序dp,每个dp[j]都在dp[j - coins[i]]上多增加了一枚硬币。
但是没有考虑到是否可取到问题,即dp[j - coins[i]]是否存在。
class Solution {
public:
int change(int amount, vector<int>& coins) {
int n = coins.size();
if (n == 0) {
return 0;
}
if (amount == 0) {
return 1;
}
vector<int> dp(amount + 1, 0);
for (int i = 0; i < n; i++) {
for (int j = amount; j >= coins[i]; j--) {
dp[j] += dp[j - coins[i]] + 1;
}
}
return dp[amount];
}
};
class Solution {
public:
int change(int amount, vector<int>& coins) {
int n = coins.size();
if (amount == 0) {
return 1;
}
vector<int> dp(amount + 1, 0);
dp[0] = 1;
for (int i = 0; i < n; i++) {
for (int j = coins[i]; j <= amount; j++) {
if (j >= coins[i]) {
dp[j] += dp[j - coins[i]];
}
}
}
return dp[amount];
}
};