4. Climbing Stairs

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You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

Leetcode: Climbing Stairs - https://leetcode.com/problems/climbing-stairs/

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解法:

Fibonacci

class Solution:
    def climbStairs(self, n: int) -> int:
        if n <= 2:
            return n
        return self.climbStairs(n-1) + self.climbStairs(n-2)

Time Limit Exceeded

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解法2: dp

O(n)

dp[i] = dp[i-1] + dp[i-2] i级台阶 = i-1 级台阶 走一级 + i-2 级台阶 走2级

class Solution:
    def climbStairs(self, n: int) -> int:
        if n <= 2:
            return n

        dp = [0] * (n+1)
        dp[1] = 1
        dp[2] = 2
        for i in range(3, n+1):
            dp[i] = dp[i-1] + dp[i-2]
        return dp[n]

[bonfy]

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].

Leetcode: 746. Min Cost Climbing Stairs - https://leetcode.com/problems/min-cost-climbing-stairs/

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解法: dp

dp[i] = min(dp[i-1], dp[i-2]) + cost[i] 表示到i级台阶的最小cost

return min(dp[-1], dp[-2]) 表示到最后一级 或者 最后第二级的 最小 cost

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        dp = [0]*(len(cost))
        dp[0], dp[1]=cost[0], cost[1]

        for i in range(2,len(cost)):
            dp[i] = min(dp[i-2]+cost[i], dp[i-1]+cost[i])

        return min(dp[-2], dp[-1])