Merge pull request #2154 from jianghongcheng/master

merge

problems/0063.不同路径II.md

@@ -397,7 +397,39 @@ class Solution:
 
 ```
 
+动态规划（版本五）
 
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid):
+        if obstacleGrid[0][0] == 1:
+            return 0
+        
+        m, n = len(obstacleGrid), len(obstacleGrid[0])
+        
+        dp = [0] * n  # 创建一个一维列表用于存储路径数
+        
+        # 初始化第一行的路径数
+        for j in range(n):
+            if obstacleGrid[0][j] == 1:
+                break
+            dp[j] = 1
+
+        # 计算其他行的路径数
+        for i in range(1, m):
+            if obstacleGrid[i][0] == 1:
+                dp[0] = 0
+            for j in range(1, n):
+                if obstacleGrid[i][j] == 1:
+                    dp[j] = 0
+                    continue
+                
+                dp[j] += dp[j - 1]
+        
+        return dp[-1]  # 返回最后一个元素，即终点的路径数
+
+
+```
 ### Go
 
 ```go

problems/0343.整数拆分.md

@@ -262,10 +262,11 @@ class Solution:
 
                 # 计算切割点j和剩余部分(i-j)的乘积，并与之前的结果进行比较取较大值
                 
-                dp[i] = max(dp[i], max((i - j) * j, dp[i - j] * j))
+                dp[i] = max(dp[i], (i - j) * j, dp[i - j] * j)
         
         return dp[n]  # 返回最终的计算结果
 
+
 ```
 动态规划（版本二）
 ```python

problems/0416.分割等和子集.md

@@ -383,6 +383,21 @@ class Solution:
             return True
         return False
 
+```
+
+卡哥版(简化版)
+```python
+class Solution:
+    def canPartition(self, nums: List[int]) -> bool:
+        if sum(nums) % 2 != 0:
+            return False
+        target = sum(nums) // 2
+        dp = [0] * (target + 1)
+        for num in nums:
+            for j in range(target, num-1, -1):
+                dp[j] = max(dp[j], dp[j-num] + num)
+        return dp[-1] == target
+
 ```
 二维DP版
 ```python

problems/0746.使用最小花费爬楼梯.md

@@ -282,7 +282,35 @@ class Solution:
         return dp1  # 返回到达楼顶的最小花费
 
 ```
+动态规划（版本三）
 
+```python
+class Solution:
+    def minCostClimbingStairs(self, cost: List[int]) -> int:
+        dp = [0] * len(cost)
+        dp[0] = cost[0]  # 第一步有花费
+        dp[1] = cost[1]
+        for i in range(2, len(cost)):
+            dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
+        # 注意最后一步可以理解为不用花费，所以取倒数第一步，第二步的最少值
+        return min(dp[-1], dp[-2])
+
+```
+动态规划（版本四）
+
+```python
+class Solution:
+    def minCostClimbingStairs(self, cost: List[int]) -> int:
+        n = len(cost)
+        prev_1 = cost[0]  # 前一步的最小花费
+        prev_2 = cost[1]  # 前两步的最小花费
+        for i in range(2, n):
+            current = min(prev_1, prev_2) + cost[i]  # 当前位置的最小花费
+            prev_1, prev_2 = prev_2, current  # 更新前一步和前两步的最小花费
+        return min(prev_1, prev_2)  # 最后一步可以理解为不用花费，取倒数第一步和第二步的最少值
+
+
+```
 ### Go
 ```Go
 func minCostClimbingStairs(cost []int) int {

problems/1049.最后一块石头的重量II.md

@@ -238,6 +238,21 @@ class Solution:
 
         return total_sum - dp[target] - dp[target]
 
+```
+
+卡哥版（简化版）
+```python
+class Solution:
+    def lastStoneWeightII(self, stones):
+        total_sum = sum(stones)
+        target = total_sum // 2
+        dp = [0] * (target + 1)
+        for stone in stones:
+            for j in range(target, stone - 1, -1):
+                dp[j] = max(dp[j], dp[j - stone] + stone)
+        return total_sum - 2* dp[-1]
+
+
 ```
 二维DP版
 ```python