@@ -305,5 +305,35 @@ int minCostClimbingStairs(int* cost, int costSize){
305305 return dp[i-1] < dp[i-2] ? dp[i-1] : dp[i-2];
306306}
307307```
308+
309+ ### Scala
310+
311+ ```scala
312+ object Solution {
313+ def minCostClimbingStairs(cost: Array[Int]): Int = {
314+ var dp = new Array[Int](cost.length)
315+ dp(0) = cost(0)
316+ dp(1) = cost(1)
317+ for (i <- 2 until cost.length) {
318+ dp(i) = math.min(dp(i - 1), dp(i - 2)) + cost(i)
319+ }
320+ math.min(dp(cost.length - 1), dp(cost.length - 2))
321+ }
322+ }
323+ ```
324+ 325+ 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] )
326+ ``` scala
327+ object Solution {
328+ def minCostClimbingStairs (cost : Array [Int ]): Int = {
329+ var dp = new Array [Int ](cost.length + 1 )
330+ for (i <- 2 until cost.length + 1 ) {
331+ dp(i) = math.min(dp(i - 1 ) + cost(i - 1 ), dp(i - 2 ) + cost(i - 2 ))
332+ }
333+ dp(cost.length)
334+ }
335+ }
336+ ```
337+ 308338-----------------------
309339<div align =" center " ><img src =https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width =500 > </img ></div >
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