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Update 0746.使用最小花费爬楼梯.md

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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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