[LeetCode] 746. Min Cost Climbing Stairs

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].

题目大意：爬楼梯，第i步将花费cost[i]，每次可以爬1格或者2格楼梯，求爬完最小的花费

分析：我们用$dp[i]$来记录爬到第i步的花费，因为每次可以爬一步或者两步，那么显然，对于最终的花费将依赖于最后两步的值，即 $dp\left[n-1\right]$ 和 $dp\left[n-2\right]$ 中的最小值，对于第i步，其最小花费为i前两步中的最小值与第i步的 $cost\left[i\right]$ 之和，即

$$dp\left[i\right]=\min\left(dp\left[i-1\right],dp\left[i-2\right]\right)+cost\left[i\right]$$ 边界条件，$dp\left[0\right]=cost\left[0\right],dp\left[1\right]=cost\left[1\right]$

时间复杂度：O(n)
空间复杂度：O(n)

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

由上面的代码可知，空间复杂度可以进一步优化为O(1)，如下

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