题解 | #跳台阶#

用函数版本

# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param number int整型 
# @return int整型
#
class Solution:
    def jumpFloor(self , number: int) -> int:
        # write code here
        #排列组合问题
        #如果是偶数，则1个2级就是n-1种情况，2个2级就是Cn-2(2)种情况，n/2个2级就是1种情况
        count = 1
        if number%2 == 0:
            for i in range(1,(number//2+1)): #i代表2级的数量
                count1 = 1
                count2 = 1
                for j in range(0,i):
                    count1 *= (number-i-j)    #A(n,k)
                    count2 *= (i-j)     #阶乘
                count += count1//count2
        else :
            for i in range(1,((number-1)//2+1)):
                count1 = 1
                count2 = 1
                for j in range(0,i):
                    count1 *= (number-i-j)    #A(n,k)
                    count2 *= (i-j)     #阶乘
                count += count1//count2
        return count
    # 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param number int整型 
# @return int整型
#
class Solution:
    
    
    def jumpFloor(self , number: int) -> int:
        # write code here
        #排列组合问题
        #如果是偶数，则1个2级就是n-1种情况，2个2级就是Cn-2(2)种情况，n/2个2级就是1种情况
        def C_function(n,k):
            count1 = 1
            count2 = 1
            for j in range(0,k):
                count1 *= (n-k-j)    #A(n,k)
                count2 *= (k-j)     #阶乘
            return (count1//count2)
    
        count = 1
        if number%2 == 0:
            for i in range(1,(number//2+1)): #i代表2级的数量
#                 count1 = 1
#                 count2 = 1
#                 for j in range(0,i):
#                     count1 *= (number-i-j)    #A(n,k)
#                     count2 *= (i-j)     #阶乘
                count += C_function(number,i)
        else :
            for i in range(1,((number-1)//2+1)):
#                 count1 = 1
#                 count2 = 1
#                 for j in range(0,i):
#                     count1 *= (number-i-j)    #A(n,k)
#                     count2 *= (i-j)     #阶乘
                count += C_function(number,i)
        return count