9. metamorphosis jump stairs
problem:
A frog can jump on a Class 1 level, you can also hop on level 2 ...... n It can also jump on stage. The frog jumped seeking a total of n grade level how many jumps.
solve:
thought:
About this problem, provided that jumps n th there will be a level of order n. analyse as below:
f(1) = 1
F (2) = F (2-1) F + (2-2) F (2-2) represents a second-order frequency hop 2 order.
f(3) = f(3-1) + f(3-2) + f(3-3)
…
f(n) = f(n-1) + f(n-2) + f(n-3) + … + f(n-(n-1)) + f(n-n)
Description:
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Where f (n) represents n-th time step has 1,2, ... n-th order of hop count method.
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n = 1, only one kind of jumps, f (1) = 1
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When n = 2, there will be two jump mode, a first-order or second-order, which return to the problem (1), f (2) = f (2-1) + f (2-2)
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For n = 3, there will be three kinds of jump mode, the first-order, second order, third order,
Then that the first stage out of the back rest 1: f (3-1); 2 out of the first order, the remaining f (3-2); 3 first order, the remaining f (3-3)
It was concluded that f (3) = f (3-1) + f (3-2) + f (3-3)
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When n = n, there are n hops manner, the first-order, second order ... n order, concluded:
f(n) = f(n-1)+f(n-2)+…+f(n-(n-1)) + f(n-n) => f(0) + f(1) + f(2) + f(3) + … + f(n-1)
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From the above is already a conclusion, but for simplicity, we can continue to simplify:
f(n-1) = f(0) + f(1)+f(2)+f(3) + … + f((n-1)-1) = f(0) + f(1) + f(2) + f(3) + … + f(n-2)
f(n) = f(0) + f(1) + f(2) + f(3) + … + f(n-2) + f(n-1) = f(n-1) + f(n-1)
It can be drawn:
f(n) = 2*f(n-1)
Then the rest is up and before the recursive + dynamic programming as a
python code:
class Solution:
def __init__(self):
self.dynamic=[None]*100
def jumpFloorII(self, number):
# write code here
result=self.dynamic[number]
if(result==None):
if(number==1):
result=1
elif(number==2):
result=2
else:
result=self.jumpFloorII(number-1)*2
self.dynamic[number]=result
return result
