1 class Solution:
 2     def coinChange(self, coins: List[int], amount: int) -> int:
 3         dp = {}
 4         dp[0] = 0  # 目标金额为0时，所需硬币数量为0
 5         for i in range(1, amount+1):  # 初始化数组
 6             dp[i] = amount+1
 7 
 8         for i in range(1, amount+1):
 9             for j in range(len(coins)):  # 每拼凑成一定金额，都来自某种面值硬币
10                 if coins[j] <= i:  # 如果当前硬币的面额可以凑
11                     dp[i] = min(dp[i], dp[i-coins[j]]+1)  # 状态转移方程：dp[i] = min{dp[i-coins[j]]}+1
12         # 如果dp[amount]的数值没有更新，说明不满足coins[j] <= i，凑不出结果
13         if dp[amount] == amount+1: return -1
14         else:   return dp[amount]

 1 class Solution {
 2 public:
 3     int coinChange(vector<int>& coins, int amount) {
 4       if(amount < 0)
 5         return -1;
 6       // 定义状态：dp[i]表示当前金额为i时至少需要dp[i]个硬币
 7       vector<int> dp(amount+1, amount+1);  // 由于将dp[0]包含进去了，所以共有amount+1个元素，并且dp元素的最大值都不会大于amount+1
 8       dp[0] = 0;
 9       for(int i = 1; i < dp.size(); i++) {
10         for(auto coin = coins.begin(); coin != coins.end(); coin++) {
11           if(i - (*coin) < 0)
12             continue;  // 由于coins的元素不一定是有序的，所以后面可能会出现符合要求的硬币
13           dp[i] = min(dp[i], dp[i-(*coin)]+1);
14         }
15       }
16       if(dp[amount] == amount + 1)
17         return -1;
18       else
19         return dp[amount];
20     }
21 };