题解 | #斐波那契数列#

# 动态规划，空间复杂度O(1)，时间复杂度O(n)
class Solution:
    def Fibonacci(self , n: int) -> int:
        if n <= 2:
            return 1
        # a和b分别为保存的前两位
        a = b = 1
        while n >= 3:
        	# tmp为当前和，更新前两位
            tmp = a + b
            a = b
            b = tmp
            n = n - 1
        return b