爬樓梯

遞歸解法：

class Solution:
    def climbStairs(self, n: int) -> int:
        cache = {}
        return self.climb(n, cache)
    def climb(self, n: int, cache):
#基本情況是0,1,2
        if n == 0:
            return 0
        elif n == 1:
            return 1
        elif n == 2:
            return 2
        elif n in cache:
            return cache[n]
        else:
            cache[n] = self.climb(n - 2, cache) + self.climb(n - 1, cache)
            return cache[n]

方法就是每次調用climb的n-2和n-1兩種情況，直到爲基本情況

斐波那契法：

n層的所有情況即爲n-1和n-2兩種，即，轉換爲斐波那契數列的計算：

class Solution:
    def climbStairs(self, n: int) -> int:
        result = [0,1,2]
        len_re = 3
        if n < len_re:
            return result[n]
        else:
            for i in range(3, n + 1):
                temp = result[-1] + result[-2]
                result[-2] = result[-1]
                result[-1] = temp
        return result[-1]