LeetCode - easy-747. Min Cost Climbing Stairs_step climb 747-优快云博客

本文链接：https://blog.youkuaiyun.com/xdczj/article/details/79212047

本文介绍了一种寻找从楼梯底部到达顶部最小成本的算法。给定每个阶梯的成本，可以选择每次爬一阶或两阶。算法提供了两种解决方案：一种使用动态规划数组，另一种则优化空间复杂度至O(1)。

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问题

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].

代码

class Solution {
public:
  //方法1：
    int minCostClimbingStairs(vector<int>& cost) {
        int N = cost.size();
        if(N == 0) return 0;
        if(N == 1) return cost[0];
        //if(N == 2) return getMin(cost[0], cost[1]);
        int dp[N+1];
        dp[0] = 0;
        dp[1] = 0;
        for(int i=2; i<=N; i++)
        {
            dp[i] = getMin(dp[i-2]+cost[i-2], dp[i-1]+cost[i-1]);
        }
        return dp[N];
    }

    int getMin(int a, int b)
    {
        return a > b ? b : a;
    }

    //methon 2: O(1) space complexity
    int minCostClimbingStairs(vector<int>& cost) {
        int N = cost.size();
        if(N == 0) return 0;
        if(N == 1) return cost[0];
        int i_2 = 0;
        int i_1 = 0;
        int temp;
        for(int i=2; i<=N; i++)
        {
            temp = i_1;
            i_1 = getMin(i_1+cost[i-1], i_2+cost[i-2]);
            i_2 = temp;
        }
        return i_1;
    }
};