算法分析与设计week13--746. Min Cost Climbing Stairs

746. Min Cost Climbing Stairs

Description

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

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].

Analysis

题意：最低成本攀登楼梯

思路：使用动态规划，递推公式：dp[i] = cost[i] + min(dp[i - 1], dp[i - 2]);

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();
        int total_cost = 0;

        if (n == 1 || n == 0) return 0;

        int cost_i_1 = 0, cost_i_2 = 0;

        for (int i = 2; i <= n; i++) {
            total_cost = min(cost[i - 1] + cost_i_1,cost[i - 2] + cost_i_2);
            cost_i_2 = cost_i_1;
            cost_i_1 = total_cost;
        }
        return total_cost;
    }
};