力扣-动态规划-518 零钱兑换Ⅱ

思路

1. dp数组定义：完全背包，不限物品使用次数，使用0-i的硬币，总和小于等于j的组合方式有dp[i][j]个
2. 递推公式：

   if(j >= coins[i])   
       dp[i][j] = dp[i-1][j] + dp[i][j - coins[i]];
   else
       dp[i][j] = dp[i-1][j];

3. dp数组初始化：第一行以及第一列初始化为1
4. 遍历顺序：左右，上下
5. 时间复杂度：     

代码

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        int m = coins.size();
        vector< vector<unsigned long long>> dp(m, vector<unsigned long long>(amount + 1, 0));

        if(coins[0] <= amount) dp[0][coins[0]] = 1;
        for (int j = coins[0] + 1; j <= amount; j++) {
            dp[0][j] = dp[0][j - coins[0]];
        }

        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }

        for(int i = 1; i < m; i++){
            for(int j = 0; j <= amount; j++){
                if(j >= coins[i])   
                    dp[i][j] = dp[i-1][j] + dp[i][j - coins[i]];
                else
                    dp[i][j] = dp[i-1][j];
            }
        }

        return dp[m-1][amount];
    }
};