746. Min Cost Climbing Stairs

xiaoxiao2021-02-28  38

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]. 题目大意：走第i步花费cost[i]，每步可以走1或2，起点可以是0或1，问走到最后的cost最小是多少。 思路： 典型的动态规划。有一个数组a来保存走到i的最少花费。那么a[i]=min(a[i-1],a[i-2])+cost[i]。最后结果便是min(a[cost.size()-1],a[cost.size()-2]). 代码： class Solution { public: int minCostClimbingStairs(vector<int>& cost) { int a[1001]; a[0]=cost[0],a[1]=cost[1]; for(int i=2;i<cost.size();i++) { a[i]=min(a[i-1],a[i-2])+cost[i]; } return min(a[cost.size()-1],a[cost.size()-2]); } };