746.min-cost-climbing-stairs
Statement
Metadata
- Link: 使用最小花费爬楼梯
- Difficulty: Easy
- Tag:
数组
动态规划
给你一个整数数组 cost
,其中 cost[i]
是从楼梯第 i
个台阶向上爬需要支付的费用。一旦你支付此费用,即可选择向上爬一个或者两个台阶。
你可以选择从下标为 0
或下标为 1
的台阶开始爬楼梯。
请你计算并返回达到楼梯顶部的最低花费。
示例 1:
输入:cost = [10,15,20]
输出:15
解释:你将从下标为 1 的台阶开始。
- 支付 15 ,向上爬两个台阶,到达楼梯顶部。
总花费为 15 。
示例 2:
输入:cost = [1,100,1,1,1,100,1,1,100,1]
输出:6
解释:你将从下标为 0 的台阶开始。
- 支付 1 ,向上爬两个台阶,到达下标为 2 的台阶。
- 支付 1 ,向上爬两个台阶,到达下标为 4 的台阶。
- 支付 1 ,向上爬两个台阶,到达下标为 6 的台阶。
- 支付 1 ,向上爬一个台阶,到达下标为 7 的台阶。
- 支付 1 ,向上爬两个台阶,到达下标为 9 的台阶。
- 支付 1 ,向上爬一个台阶,到达楼梯顶部。
总花费为 6 。
提示:
2 <= cost.length <= 1000
0 <= cost[i] <= 999
Metadata
- Link: Min Cost Climbing Stairs
- Difficulty: Easy
- Tag:
Array
Dynamic Programming
You are given an integer array cost
where cost[i]
is the cost of ith
step on a staircase. Once you pay the cost, you can either climb one or two steps.
You can either start from the step with index 0
, or the step with index 1
.
Return the minimum cost to reach the top of the floor.
Example 1:
Input: cost = [10,15,20]
Output: 15
Explanation: You will start at index 1.
- Pay 15 and climb two steps to reach the top.
The total cost is 15.
Example 2:
Input: cost = [1,100,1,1,1,100,1,1,100,1]
Output: 6
Explanation: You will start at index 0.
- Pay 1 and climb two steps to reach index 2.
- Pay 1 and climb two steps to reach index 4.
- Pay 1 and climb two steps to reach index 6.
- Pay 1 and climb one step to reach index 7.
- Pay 1 and climb two steps to reach index 9.
- Pay 1 and climb one step to reach the top.
The total cost is 6.
Constraints:
2 <= cost.length <= 1000
0 <= cost[i] <= 999
Solution
from typing import List
class Solution:
def minCostClimbingStairs(self, cost: List[int]) -> int:
cost.insert(0, 0)
cost.append(0)
n = len(cost)
f = [0x3f3f3f3f for i in range(n)]
f[0] = 0
for i in range(n):
if i + 1 < n:
f[i + 1] = min(f[i + 1], f[i] + cost[i])
if i + 2 < n:
f[i + 2] = min(f[i + 2], f[i] + cost[i])
return f[-1]
最后更新: October 11, 2023