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Dynamic Programming - 70. Climbing Stairs

Dynamic Programming - 70. Climbing Stairs

70. Climbing Stairs

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

思路:

爬楼梯和62题属于典型的斐波那契数列型的动态规划。做动态规划题目主要是四步:

  1. 确定状态,找出问题中的子问题,考虑最后一步是什么。
  2. 写出状态转移方程。
  3. 定义初始条件和边界条件(Corner Case)
  4. 计算顺序,也就是从小到大还是从大到小,爬楼梯问题很明显,是从小到大,备忘录类型。
    这题中,我的dp[i]代表的是到达第i-1台台阶需要走几步,所以dp[0]代表的是第一台台阶需要的步数,所以我的slice长度开辟了n。

代码:

go:

func climbStairs(n int) int {

    if n <= 0 {
        return 0
    }
    
    if (n == 1 || n == 2) {
        return n
    }
    dp := make([]int, n)
    // initialization
    dp[0] = 1
    dp[1] = 2
    
    for i := 2; i < n; i++ {
        dp[i] = dp[i-1] + dp[i-2]
    }
    
    return dp[n-1]
}


标题:Dynamic Programming - 70. Climbing Stairs
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/07/31/1564582688150.html

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