746. Min Cost Climbing Stairs

xiaoxiao2021-02-28  13

On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).

Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1.

Example 1:

Input: cost = [10, 15, 20] Output: 15 Explanation: Cheapest is start on cost[1], pay that cost and go to the top.

Example 2:

Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1] Output: 6 Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3].

Note:

cost will have a length in the range [2, 1000].Every cost[i] will be an integer in the range [0, 999].

类似于斐波数列,简单的DP。

class Solution { public int minCostClimbingStairs(int[] cost) { if(cost.length == 1) return cost[0]; if(cost.length == 2) return Math.min(cost[0],cost[1]); for(int i = 2;i<cost.length;i++){ cost[i] = Math.min(cost[i-1],cost[i-2])+cost[i]; } return Math.min(cost[cost.length-1],cost[cost.length-2]); } }

看到好多AC代码没有判断长度,无所关系。

要注意的点就是最后不能直接返回数组最后一个值,因为可能两步跳出,如例一。

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