参考代码：

未优化代码：

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
        const int INF = 0x3f3f3f3f;
        //多开一行，多开一列
        vector<vector<int>> dp(n + 1, vector<int>(amount + 1));

        //初始化
        dp[0][0] = 0;
        for (int j = 1; j <= amount; j++)
        {
            //根据后面填表时取min的性质，所以无效值应该设置成正无穷大
            dp[0][j] = INF;
        }

        //填表
        for (int i = 1; i <= n; i++)
        {
            for (int j = 0; j <= amount; j++)
            {
                dp[i][j]=dp[i - 1][j];
                if(j>=coins[i-1])
                {
                    //注意，这里是取min，所以不存在的值应该设成正无穷大才对，不能选择-1作为无效值
                    dp[i][j]=min(dp[i][j],dp[i][j - coins[i - 1]]+1);
                }
            }
        }

        return dp[n][amount]>=INF?-1:dp[n][amount];
    }
};

优化后的代码：

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
        const int INF = 0x3f3f3f3f;
        //多开一行，多开一列
        //初始化
        vector<int> dp(amount + 1,INF);
        dp[0] = 0;

        //填表
        for (int i = 1; i <= n; i++)
        {
            for (int j = coins[i-1]; j <= amount; j++)
            {
                //注意，这里是取min，所以不存在的值应该设成正无穷大才对，不能选择-1作为无效值
                dp[j]=min(dp[j],dp[j - coins[i - 1]]+1);
            }
        }

        return dp[amount]>=INF?-1:dp[amount];
    }
};

你学会了吗？？？