lintcode740. 零钱兑换 2 完全揹包

给出不同面值的硬币以及总金额. 试写一函数来计算构成该总额的组合数量. 你可以假设每一种硬币你都有无限个.

样例
样例1

输入: amount = 10 和 coins = [10] 
输出: 1
样例2

输入: amount = 8 和 coins = [2, 3, 8]
输出: 3
解释:
有3种方法:
8 = 8
8 = 3 + 3 + 2
8 = 2 + 2 + 2 + 2
注意事项
你可以做出以下假设:

0 <= amount <= 5000

1 <= coin <= 5000

硬币种类不超过 500

结果保证符合 32 位符号整数

class Solution {
public:
    /**
     * @param amount: a total amount of money amount
     * @param coins: the denomination of each coin
     * @return: the number of combinations that make up the amount
     */
    int change(int amount, vector<int> &coins) {
        // write your code here
        vector<int>dp(amount+1,0);
        dp[0]=1;
        for (int i = 0; i < coins.size(); i++) {
            for(int j = coins[i];j<=amount;j++)
            {
                dp[j]+=dp[j-coins[i]];
            }
        }
        return dp[amount];
    }
};