LeetCode 746. Min Cost Climbing Stairs

原题链接在这里：https://leetcode.com/problems/min-cost-climbing-stairs/description/

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]

.

题解：

f(n)代表走到n需要的最小花费. f(n) = Math.min(f(n-1), f(n-1)) + cost
.

跳到最后可能跳两步直接跳到最后, 所以要在 用这一步 和 这一步前一步 之间选出最小值.

Time Complexity: O(cost.length).

Space: O(1).

AC Java:

1 class Solution {
2     public int minCostClimbingStairs(int[] cost) {
3         if(cost == null || cost.length == 0){
4             return 0;
5         }
6
7         int first = 0;
8         int second = 0;
9         int third = 0;
10         for(int i = 0; i<cost.length; i++){
11             third = Math.min(first, second) + cost[i];
12             first = second;
13             second = third;
14         }
15
16         return Math.min(first, second);
17     }
18 }

类似Climbing Stairs.