Min Cost Climbing Stairs

发布时间：2020-12-14 04:49:38 所属栏目：大数据来源：网络整理

导读：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 fro

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]
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].

 1 class Solution {
 2     public int minCostClimbingStairs(int[] cost) {
 3         int n = cost.length;
 4         int num1 = cost[0],num2 = cost[1];
 5         
 6         for (int i = 2; i < n; i++) {
 7             int temp = Math.min(num1,num2) + cost[i];
 8             num1 = num2;
 9             num2 = temp;
10         }
11         
12         return Math.min(num1,num2);
13     }
14 }

 1 class Solution {
 2     public int minCostClimbingStairs(int[] cost) {
 3         int n = cost.length;
 4         int minCost[] = new int[n];
 5         minCost[0] = cost[0];
 6         minCost[1] = cost[1];
 7         
 8         for (int i = 2; i < n; i++) {
 9             minCost[i] = Math.min(minCost[i - 2],minCost[i - 1]) + cost[i];
10         }
11         
12         return Math.min(minCost[n - 2],minCost[n - 1]);
13     }
14 }

（编辑：李大同）

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