【easy】746. Min Cost Climbing Stairs 动态规划

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:

1. cost will have a length in the range [2, 1000].
2. Every cost[i] will be an integer in the range [0, 999].

方法一：正序

/**
     * @param cost 每一步所要花费的值
     * @return 到达顶部总共需要的值
     */
    public int minCostClimbingStairs(int[] cost) {
        int length = cost.length + 1;
        int[] dp = new int[length];
        dp[0] = 0;
        dp[1] = 0;
        for (int i = 2; i < length; i++) {
            dp[i] = Math.min(dp[i - 2] + cost[i - 2], dp[i - 1] + cost[i - 1]);
        }
        return dp[length - 1];
    }

方法二：倒序

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int f1 = 0, f2 = 0;
        for (int i=cost.size()-1; i>= 0; i--){
            int f0 = cost[i] + min(f1, f2);
            f2 = f1;
            f1 = f0;
        }
        return min(f1, f2);
    }
};

【easy】746. Min Cost Climbing Stairs 动态规划

原文：https://www.cnblogs.com/sherry-yang/p/10772154.html