添加 0746.使用最小花费爬楼梯.md Scala版本

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ZongqinWang
2022-06-18 20:12:19 +08:00
parent 5985aae83c
commit 6e92cd2417

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@ -305,5 +305,35 @@ int minCostClimbingStairs(int* cost, int costSize){
return dp[i-1] < dp[i-2] ? dp[i-1] : dp[i-2];
}
```
### Scala
```scala
object Solution {
def minCostClimbingStairs(cost: Array[Int]): Int = {
var dp = new Array[Int](cost.length)
dp(0) = cost(0)
dp(1) = cost(1)
for (i <- 2 until cost.length) {
dp(i) = math.min(dp(i - 1), dp(i - 2)) + cost(i)
}
math.min(dp(cost.length - 1), dp(cost.length - 2))
}
}
```
第二种思路: dp[i] 表示爬到第i-1层所需的最小花费状态转移方程为: dp[i] = min(dp[i-1]+cost[i-1],dp[i-2]+cost[i-2])
```scala
object Solution {
def minCostClimbingStairs(cost: Array[Int]): Int = {
var dp = new Array[Int](cost.length + 1)
for (i <- 2 until cost.length + 1) {
dp(i) = math.min(dp(i - 1) + cost(i - 1), dp(i - 2) + cost(i - 2))
}
dp(cost.length)
}
}
```
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>