[pull] master from youngyangyang04:master #293

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Expand Up		@@ -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
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3 changes: 2 additions & 1 deletion problems/0343.整数拆分.md

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Expand Up		@@ -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
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15 changes: 15 additions & 0 deletions problems/0416.分割等和子集.md

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Expand Up		@@ -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
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28 changes: 28 additions & 0 deletions problems/0746.使用最小花费爬楼梯.md

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Expand Up		@@ -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 {
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15 changes: 15 additions & 0 deletions problems/1049.最后一块石头的重量II.md

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Expand Up		@@ -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
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