feat: add Python Code

problems/518.coin-change-2.md

@@ -117,6 +117,12 @@ return dp[dp.length - 1];
 
 ## 代码
 
+
+代码支持：Python3，JavaScript：
+
+
+JavaSCript Code:
+
 ```js
 /*
  * @lc app=leetcode id=518 lang=javascript
@@ -146,10 +152,46 @@ var change = function(amount, coins) {
 };
 ```
 
+Python Code:
+
+```
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        dp = [0] * (amount + 1)
+        dp[0] = 1
+
+        for j in range(len(coins)):
+            for i in range(1, amount + 1):
+                if i >= coins[j]:
+                    dp[i] += dp[i - coins[j]]
+
+        return 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。
 
 相似问题:
 
 [322.coin-change](./322.coin-change.md)
+
+## 附录
+
+Python 二维解法（不推荐）：
+
+```python
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        dp = [[0 for _ in range(len(coins) + 1)] for _ in range(amount + 1)]
+        for j in range(len(coins) + 1):
+            dp[0][j] = 1
+
+        for i in range(amount + 1):
+            for j in range(1, len(coins) + 1):
+                if i >= coins[j - 1]:
+                    dp[i][j] = dp[i - coins[j - 1]][j] + dp[i][j - 1]
+                else:
+                    dp[i][j] = dp[i][j - 1]
+        return dp[-1][-1]
+```