LeetCode-746. Min Cost Climbing Stairs

LeetCode-746. Min Cost Climbing Stairs

​​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: 15Explanation: 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: 6Explanation: 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]​​.

题解：动态规划，每次把0-i的最小代价算出然后动态更新。

class Solution {public: int minCostClimbingStairs(vector& cost) { cost.push_back(0); int n = cost.size(); if (n == 1) { return 0; } vector dp(n, 0); dp[0] = cost[0]; dp[1] = cost[1]; for (int i = 2; i < n; i++) { dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]; } return dp[n - 1]; }};

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