518. Coin Change 2

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.

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

Note:

You can assume that

1. 0 <= amount <= 5000
2. 1 <= coin <= 5000
3. the number of coins is less than 500
4. the answer is guaranteed to fit into signed 32-bit integer

方法1: 2d-dynamic programming

class Solution1 {
public:
    int change(int amount, vector<int>& coins) {
        int n = coins.size();
        vector<vector<int>> dp(n + 1, vector<int>(amount + 1, 0));
        for (int i = 0; i < n + 1; i++) dp[i][0] = 1;
        for (int i = 1; i < n + 1; i++) {
            for (int j = 1; j < amount + 1; j++) {
                if (j - coins[i - 1] >= 0) dp[i][j] += dp[i][j - coins[i - 1]];
                dp[i][j] += dp[i - 1][j];
            }
        }
        return dp.back().back();
    }
};

方法2: 1d-dynamic programming

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        int n = coins.size();
        vector<int> dp(amount + 1, 0);
        dp[0] = 1;
        for (int i = 0; i < n; i++) {
            for (int j = coins[i]; j < amount + 1; j++) {
                if (j - coins[i] >= 0) dp[j] += dp[j - coins[i]]; 
            }
        }
        return dp.back();
    }
};