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Merge pull request youngyangyang04#2154 from jianghongcheng/master

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`@@ -397,7 +397,39 @@ class Solution:`
`397`	`397`
`398`	`398`	```
`399`	`399`
	`400`	`+动态规划(版本五)`
`400`	`401`
	`402`	+```python
	`403`	`+class Solution:`
	`404`	`+ def uniquePathsWithObstacles(self, obstacleGrid):`
	`405`	`+ if obstacleGrid[0][0] == 1:`
	`406`	`+ return 0`
	`407`	`+`
	`408`	`+ m, n = len(obstacleGrid), len(obstacleGrid[0])`
	`409`	`+`
	`410`	`+ dp = [0] * n # 创建一个一维列表用于存储路径数`
	`411`	`+`
	`412`	`+ # 初始化第一行的路径数`
	`413`	`+ for j in range(n):`
	`414`	`+ if obstacleGrid[0][j] == 1:`
	`415`	`+ break`
	`416`	`+ dp[j] = 1`
	`417`	`+`
	`418`	`+ # 计算其他行的路径数`
	`419`	`+ for i in range(1, m):`
	`420`	`+ if obstacleGrid[i][0] == 1:`
	`421`	`+ dp[0] = 0`
	`422`	`+ for j in range(1, n):`
	`423`	`+ if obstacleGrid[i][j] == 1:`
	`424`	`+ dp[j] = 0`
	`425`	`+ continue`
	`426`	`+`
	`427`	`+ dp[j] += dp[j - 1]`
	`428`	`+`
	`429`	`+ return dp[-1] # 返回最后一个元素,即终点的路径数`
	`430`	`+`
	`431`	`+`
	`432`	+```
`401`	`433`	`### Go`
`402`	`434`
`403`	`435`	```go

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`@@ -262,10 +262,11 @@ class Solution:`
`262`	`262`
`263`	`263`	`# 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`
`264`	`264`
`265`		`- dp[i] = max(dp[i], max((i - j) * j, dp[i - j] * j))`
	`265`	`+ dp[i] = max(dp[i], (i - j) * j, dp[i - j] * j)`
`266`	`266`
`267`	`267`	`return dp[n] # 返回最终的计算结果`
`268`	`268`
	`269`	`+`
`269`	`270`	```
`270`	`271`	`动态规划(版本二)`
`271`	`272`	```python

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`@@ -383,6 +383,21 @@ class Solution:`
`383`	`383`	`return True`
`384`	`384`	`return False`
`385`	`385`
	`386`	+```
	`387`	`+`
	`388`	`+卡哥版(简化版)`
	`389`	+```python
	`390`	`+class Solution:`
	`391`	`+ def canPartition(self, nums: List[int]) -> bool:`
	`392`	`+ if sum(nums) % 2 != 0:`
	`393`	`+ return False`
	`394`	`+ target = sum(nums) // 2`
	`395`	`+ dp = [0] * (target + 1)`
	`396`	`+ for num in nums:`
	`397`	`+ for j in range(target, num-1, -1):`
	`398`	`+ dp[j] = max(dp[j], dp[j-num] + num)`
	`399`	`+ return dp[-1] == target`
	`400`	`+`
`386`	`401`	```
`387`	`402`	`二维DP版`
`388`	`403`	```python

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`@@ -282,7 +282,35 @@ class Solution:`
`282`	`282`	`return dp1 # 返回到达楼顶的最小花费`
`283`	`283`
`284`	`284`	```
	`285`	`+动态规划(版本三)`
`285`	`286`
	`287`	+```python
	`288`	`+class Solution:`
	`289`	`+ def minCostClimbingStairs(self, cost: List[int]) -> int:`
	`290`	`+ dp = [0] * len(cost)`
	`291`	`+ dp[0] = cost[0] # 第一步有花费`
	`292`	`+ dp[1] = cost[1]`
	`293`	`+ for i in range(2, len(cost)):`
	`294`	`+ dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]`
	`295`	`+ # 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`
	`296`	`+ return min(dp[-1], dp[-2])`
	`297`	`+`
	`298`	+```
	`299`	`+动态规划(版本四)`
	`300`	`+`
	`301`	+```python
	`302`	`+class Solution:`
	`303`	`+ def minCostClimbingStairs(self, cost: List[int]) -> int:`
	`304`	`+ n = len(cost)`
	`305`	`+ prev_1 = cost[0] # 前一步的最小花费`
	`306`	`+ prev_2 = cost[1] # 前两步的最小花费`
	`307`	`+ for i in range(2, n):`
	`308`	`+ current = min(prev_1, prev_2) + cost[i] # 当前位置的最小花费`
	`309`	`+ prev_1, prev_2 = prev_2, current # 更新前一步和前两步的最小花费`
	`310`	`+ 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`
	`311`	`+`
	`312`	`+`
	`313`	+```
`286`	`314`	`### Go`
`287`	`315`	```Go
`288`	`316`	`func minCostClimbingStairs(cost []int) int {`

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`@@ -238,6 +238,21 @@ class Solution:`
`238`	`238`
`239`	`239`	`return total_sum - dp[target] - dp[target]`
`240`	`240`
	`241`	+```
	`242`	`+`
	`243`	`+卡哥版(简化版)`
	`244`	+```python
	`245`	`+class Solution:`
	`246`	`+ def lastStoneWeightII(self, stones):`
	`247`	`+ total_sum = sum(stones)`
	`248`	`+ target = total_sum // 2`
	`249`	`+ dp = [0] * (target + 1)`
	`250`	`+ for stone in stones:`
	`251`	`+ for j in range(target, stone - 1, -1):`
	`252`	`+ dp[j] = max(dp[j], dp[j - stone] + stone)`
	`253`	`+ return total_sum - 2* dp[-1]`
	`254`	`+`
	`255`	`+`
`241`	`256`	```
`242`	`257`	`二维DP版`
`243`	`258`	```python

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