Merge pull request #1873 from Logenleedev/my-contribution

添加 斐波那契数动态规划 python 版本 方便理解

problems/0416.分割等和子集.md

@@ -291,18 +291,63 @@ false true false false false true true false false false true true
 
 ### Python：
 ```python
+# 一维度数组解法
 class Solution:
     def canPartition(self, nums: List[int]) -> bool:
         target = sum(nums)
         if target % 2 == 1: return False
         target //= 2
-        dp = [0] * 10001
+        dp = [0] * (len(nums) + 1)
         for i in range(len(nums)):
             for j in range(target, nums[i] - 1, -1):
                 dp[j] = max(dp[j], dp[j - nums[i]] + nums[i])
         return target == dp[target]
 ```
 
+```python
+# 二维度数组解法
+class Solution:
+    def canPartition(self, nums: List[int]) -> bool:
+        target = sum(nums)
+        nums = sorted(nums)
+
+        # 做最初的判断
+        if target % 2 != 0:
+            return False
+
+        # 找到 target value 可以认为这个是背包的体积
+        target = target // 2
+
+        row = len(nums)
+        col = target + 1
+
+        # 定义 dp table
+        dp = [[0 for _ in range(col)] for _ in range(row)] 
+
+        # 初始 dp value
+        for i in range(row):
+            dp[i][0] = 0
+        
+        for j in range(1, target):
+            if nums[0] <= j:
+                dp[0][j] = nums[0]
+
+        # 遍历 先遍历物品再遍历背包
+        for i in range(1, row):
+
+            cur_weight = nums[i]
+            cur_value = nums[i]
+
+            for j in range(1, col):
+                if cur_weight > j:
+                    dp[i][j] = dp[i - 1][j]
+                else:
+                    dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - cur_weight] + cur_value)
+            
+        # 输出结果
+        return dp[-1][col - 1] == target   
+```
+
 ### Go：
 ```go
 // 分割等和子集 动态规划

problems/0509.斐波那契数.md

@@ -214,6 +214,33 @@ class Solution:
             a, b = b, c
         return c
 
+# 动态规划 (注释版。无修饰)
+class Solution:
+    def fib(self, n: int) -> int:
+       
+        # 排除 Corner Case
+        if n == 1:
+            return 1
+
+        if n == 0:
+            return 0
+        
+        # 创建 dp table 
+        dp = [0] * (n + 1)
+
+        # 初始化 dp 数组
+        dp[0] = 0
+        dp[1] = 1
+
+        # 遍历顺序: 由前向后。因为后面要用到前面的状态
+        for i in range(2, n + 1):
+
+            # 确定递归公式/状态转移公式
+            dp[i] = dp[i - 1] + dp[i - 2]
+        
+        # 返回答案
+        return dp[n]
+
 # 递归实现
 class Solution:
     def fib(self, n: int) -> int:

problems/1049.最后一块石头的重量II.md

@@ -224,7 +224,7 @@ class Solution:
     def lastStoneWeightII(self, stones: List[int]) -> int:
         sumweight = sum(stones)
         target = sumweight // 2
-        dp = [0] * 15001
+        dp = [0] * (target + 1)
         for i in range(len(stones)):
             for j in range(target, stones[i] - 1, -1):
                 dp[j] = max(dp[j], dp[j - stones[i]] + stones[i])