[lintcode]740. 零钱兑换II

链接: http://www.lintcode.com/zh-cn/problem/coin-change-ii/#

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.

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) {
        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];
    }
};