From c736cffe22143965ffc50896ca5b11a3e65df536 Mon Sep 17 00:00:00 2001
From: zhqiao <qbuer.qbuer@gmail.com>
Date: Tue, 2 Jan 2024 21:16:29 +0800
Subject: [PATCH] =?UTF-8?q?0377=20=E6=96=B0=E5=A2=9Epython=20=E4=BA=8C?=
 =?UTF-8?q?=E7=BB=B4DP=E6=95=B0=E7=BB=84=E7=89=88?=
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---
 problems/0377.组合总和Ⅳ.md | 28 ++++++++++++++++++++++++++++
 1 file changed, 28 insertions(+)

diff --git a/problems/0377.组合总和Ⅳ.md b/problems/0377.组合总和Ⅳ.md
index d9699c54..05f852b1 100644
--- a/problems/0377.组合总和Ⅳ.md
+++ b/problems/0377.组合总和Ⅳ.md
@@ -206,6 +206,34 @@ class Solution:
 
 
 ```
+二维DP版
+```python
+class Solution:
+    def combinationSum4(self, nums: List[int], target: int) -> int:
+        # dp[][j]和为j的组合的总数
+        dp = [[0] * (target+1) for _ in nums]
+        
+        for i in range(len(nums)):
+            dp[i][0] = 1
+            
+        # 这里不能初始化dp[0][j]。dp[0][j]的值依赖于dp[-1][j-nums[0]]
+            
+        for j in range(1, target+1):
+            for i in range(len(nums)):
+                
+                if j - nums[i] >= 0:
+                    dp[i][j] = (
+                        # 不放nums[i]
+                        # i = 0 时，dp[-1][j]恰好为0，所以没有特殊处理
+                        dp[i-1][j] +
+                        # 放nums[i]。对于和为j的组合，只有试过全部物品，才能知道有几种组合方式。所以取最后一个物品dp[-1][j-nums[i]]
+                        dp[-1][j-nums[i]]
+                    )
+                else:
+                    dp[i][j] = dp[i-1][j]
+        return dp[-1][-1]
+```
+
 ### Go：
 
 ```go