LetCode 70. Climbing Stairs--动态规划-爬梯子--递归等解法

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LeetCode 动态规划（Dynamic programming）系列题目:LeetCode 动态规划（Dynamic programming）系列题目

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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

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爬梯子，动态规划即可。
Python代码如下：

class Solution:
    def climbStairs(self, n: int) -> int:
        t=[1,2]
        for i in range(2,n):
            t.append(t[i-1]+t[i-2])
        return t[n-1]

java解法：

class Solution {
    public int climbStairs(int n) {
        int []dp=new int[n+1];
        if(n==1){
            return 1;
        }
        dp[1]=1;
        dp[2]=2;
        for (int i=3;i<=n;i++){
            dp[i]=dp[i-1]+dp[i-2];
        }
        return dp[n];
    }
}

python如果直接使用递归的话，会爆内存，使用缓存可以解决这个问题：

import functools

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

可以直接使用斐波那契公式，Java代码如下：

public class Solution {
    public int climbStairs(int n) {
        double sqrt5=Math.sqrt(5);
        double fibn=Math.pow((1+sqrt5)/2,n+1)-Math.pow((1-sqrt5)/2,n+1);
        return (int)(fibn/sqrt5);
    }
}