LeetCode 746. Min Cost Climbing Stairs

[Woodyiiiiiii]

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].

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Climbing Stairs的扩展，利用dp数组存储路径。
可以这样写：

//Solution1:
class Solution {
public int minCostClimbingStairs(int[] cost) {
        int len = cost.length;
        int[] dp = new int[len + 1];
        for (int i = 2; i < len + 1; ++i) {
            dp[i] = Math.min(cost[i - 2] + dp[i - 2], cost[i - 1] + dp[i - 1]);
        }
        
        return dp[len];
    }
}

也可以这样写。

//Solution2:
class Solution {
public int minCostClimbingStairs(int[] cost) {
        int len = cost.length;
        int[] dp = new int[len];
        dp[0] = cost[0];
        dp[1] = cost[1];
        for (int i = 2; i < len; ++i) {
            dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
        }
        
        return Math.min(dp[len - 1], dp[len - 2]);
    }
}

可以继续优化，实际上只需要两个变量就可以了，将空间复杂度减少至O(1)。

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int a = 0;
        int b = 0;
        for (int c : cost) {
            int t = Math.min(a, b) + c;
            a = b;
            b = t;
        }
        
        return Math.min(a, b);
    }
}

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参考资料:

1. LeetCode原题
2. grandyang题解