746. Min Cost Climbing Stairs

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本文介绍了一种使用动态规划解决的算法问题——寻找到达楼层顶部的最小成本路径。该问题设定为在一个楼梯上，每一步都有一个非负的成本，你可以选择一步或两步前进。从第0步或第1步开始，目标是最小化到达顶部的成本。

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题目

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].

思路

动态规划的思想。假如[a,b,c]那么a=min(a+b,a+c)。

代码

class Solution {
public:
    int minCostClimbingStairs(vector<int> cost) {
        int result=0;
        if(cost.size()==2) {
            return cost[0]>cost[1]?cost[1]:cost[0];
        }
        for(int i=cost.size()-3;i>=0;i--) {
            cost[i]=(cost[i]+cost[i+1])>(cost[i]+cost[i+2])?(cost[i]+cost[i+2]):(cost[i]+cost[i+1]);
        }
        return cost[0]>cost[1]?cost[1]:cost[0];
    }
};