518. 零钱兑换 II

完全背包

选法并不会重复，

因为是每个单独，只用一次

class Solution {
public:
    int dp[5100];
    int change(int amount, vector<int>& coins) {
        memset(dp, 0, sizeof(dp));
        dp[0] = 1;

        for(int i = 1; i <= coins.size(); i++)
        {
            int coin = coins[i - 1];
            for(int j = coin; j <= amount; j++)
            {
                dp[j] += dp[j - coin];
            }
        }
        return dp[amount];
        




    }
};