[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

• 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

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

思路：动规问题的核心就是状态递归方程，本题为f(n)=f(n-coin[0])+f(n-coin[1])+...+f(n-coin[n])(假设n>coin[n]);