思路：
https://www.youtube.com/watch?v=uUETHdijzkA

方法1: DP, 2-d 数组

这种方法浪费空间

1. 易错点：第一行初始值要设为amount + 1, i.e. 一个比任何可能解都更大的数。之所以不取INT_MAX是为了防止出现INT_MAX + 1造成的overflow
2. 两个向量唯独都是 n + 1, m + 1
3. 但是coin和dp的纵向坐标相差1，不处理会出现heap-overflow

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
        
        vector<vector<int>> dp(n + 1, vector<int>(amount + 1, amount + 1));
        for (int i = 0; i < n + 1; i ++){
            dp[i][0] = 0;     
        }
       
        for (int i = 1; i < n + 1; i ++){
            for (int j = 1; j < amount + 1; j ++){
                if (j - coins[i- 1] >= 0){
                    dp[i][j] = min(dp[i - 1][j], dp[i][j - coins[i - 1]] + 1);
                } 
                else
                    dp[i][j] = dp[i - 1][j];
                }
            }
        // printMatrix(dp);
        return dp[n][amount] > amount ? -1 : dp[n][amount];
    }
private: 
    void printMatrix(vector<vector<int>> & A){
        for (int i = 0; i < A.size(); i++){
            for (int j = 0; j < A[0].size() ; j ++){
                cout << A[i][j] << " ";
            }
            cout << endl;
        }
    }
};

方法2: DP，1-d 数组

每次新一行都迭代上一行，不需要建立2-d数组

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int Max = amount + 1;
        vector<int> dp(amount + 1, Max);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int j = 0; j < coins.size(); j++) {
                if (coins[j] <= i) {
                    dp[i] = min(dp[i], dp[i - coins[j]] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
};

小改动

 int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
        vector<int> dp(amount + 1, amount + 1);
        dp[0] = 0;
        
        for (int coin: coins){
            for (int i = coin; i <= amount; i++ )
                dp[i] = min(dp[i], dp[i - coin] + 1);
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
    

方法3: 剪枝