feat: add python code

Author: azl397985856

problems/322.coin-change.md

@@ -73,7 +73,8 @@ eg: 对于 [1,2,5] 组成 11 块
 
 ## 代码
 
-* 语言支持：JS，C++
+
+* 语言支持：JS，C++,Python3
 
 JavaScript Code：
 ```js
@@ -116,6 +117,51 @@ public:
     }
 };
 ```
+
+Python3 Code:
+
+
+(二维数组)
+
+```python
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        if amount < 0:
+            return - 1
+        dp = [[amount + 1 for _ in range(len(coins) + 1)]
+              for _ in range(amount + 1)]
+        # 初始化第一行为0，其他为最大值（也就是amount + 1）
+
+        for j in range(len(coins) + 1):
+            dp[0][j] = 0
+
+        for i in range(1, amount + 1):
+            for j in range(1, len(coins) + 1):
+                if i - coins[j - 1] >= 0:
+                    dp[i][j] = min(
+                        dp[i][j - 1], dp[i - coins[j - 1]][j] + 1)
+                else:
+                    dp[i][j] = dp[i][j - 1]
+
+        return -1 if dp[-1][-1] == amount + 1 else dp[-1][-1]
+ ```
+ 
+ 
+ （一维数组）
+ 
+```python
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        dp = [amount + 1] * (amount + 1)
+        dp[0] = 0
+
+        for i in range(1, amount + 1):
+            for j in range(len(coins)):
+                if i >= coins[j]:
+                    dp[i] = min(dp[i], dp[i - coins[j]] + 1)
+
+        return -1 if dp[-1] == amount + 1 else dp[-1]
+```
 ## 扩展
 
 这是一道很简单描述的题目， 因此很多时候会被用到大公司的电面中。