Three methods for jumping step problem (abnormal jumping steps)

Topic 1: A step of a total of n-level, if you can jump 1 levels, you can jump 2 levels. How many total hops are in total, and the time complexity of the algorithm is analyzed.

We think of the N-Step jumping method as a function of N, which is recorded as f (n). When N>2, the first jump when there are two different choices: first, only jump 1, when the number of hops is equal to the rest of the n-1 steps of the number of jumps, that is f (n-1); Another option is the first jump 2, at this time the number of hops is equal to the rest of the n-2 step of the number of jumps, That is f (n-2). Thus the total number of different hops for N-Step steps f (n) =f (n-1) + (f-2).

We summarize the above analysis with a formula as follows:

/1 N=1
F (n) = 2 n=2
\ f (n-1) + (f-2) n>2

Analysis here, I believe a lot of people can see that this is our familiar Fibonacci sequence.

Here are my three ways to solve this:

#include <stdio.h>
#include <string.h>
Recursive solution
int solution1 (int n)
{
if (n = = 0 | | n = = 1) return 1;
else return Solution1 (n-1) + solution1 (n-2);
}
Non-recursive solution
int Solution2 (int n)
{
int f[100];
F[0] = 1;
F[1] = 1;
for (int i=2; i<=n; ++i)
F[i] = F[i-1] + f[i-2];
return f[n];
}
Method of scrolling Array
int Solution3 (int n)
{
int f[3]={0,1,2};
if (n<=2)
{
return f[n];
}
for (int i=3;i<=n;i++)
{
F[0]=F[1];
F[1]=F[2];
F[2]=F[0]+F[1];
}
return f[2];
}
int main ()
{
int n;
printf ("Please enter the number of steps:");
scanf ("%d", &n);
int sum1= solution1 (n);
int sum2= solution2 (n);
int sum3= solution3 (n);
printf ("sum1=%d,sum2=%d,sum3=%d\n", sum1,sum2,sum3);
return 0;
}

Topic 2: A step has a total of n, if you can jump 1 levels at a time, you can jump 2 levels ... It can also jump on n levels. At this point, the frog jumps to the level of the N-step of the total number of jumping method?
 Analysis: Using FIB (n) to indicate the frog jumps on the N-Step steps of the number of hops, the frog jumped on the N-Step step of the number of hops 1 (n-Step), set fib (0) = 1;
when n = 1 o'clock, there is only one method of jumping, that is, the 1-Step Jump: Fib (1) = 1;
when n = 2 o'clock, there are two ways of jumping, one-step and second-order jumps: fib (2) = FIB (1) + fib (0) = 2;
when n = 3 o'clock, there are three ways to jump, the first step out of one, followed by FIB (3-1) in the Jump method, the first second out of the second, there is a fib (3-2) in the Jump method, after the first jump out of third order, followed by FIB (3-3) in the Jump method
fib (3) = FIB (2) + fib (1) +fib (0) =4;
when n = N, there is a total of n jumps, the first step out of one order, followed by FIB (n-1) in the Jump method, after the first second out of the second, there is a fib (n-2) in the Jump method ..... ..... After the first step out of the N-order, there is a Fib (n-n) jump in the back.
fib (n) = fib (n-1) +fib (n-2) +fib (n-3) +..........+fib (n-n) =fib (0) +fib (1) +fib (2) +.......+fib (n-1)
and because Fib (n-1) =fib (0) +fib (1) +fib (2) +.......+fib (n-2)
two-type subtraction: Fib (n)-fib (n-1) =fib (n-1) ===== "Fib (n) = 2*fib (n-1)N >= 2
The recursive equation is as follows:

Freak Jump
int solution4 (int n)
{
if (n = = 0 | | n = = 1)
return 1;
Else
Return 2*solution4 (n-1);
}

Perverted jump Scrolling Array
int solution5 (int n)
{
int f[2]={0,1};
if (n < 2)
return f[n];

for (int i=2;i<=n;i++)
{
F[0]=F[1];
F[1]=2*F[0];
}
return f[1];
}

Three methods for jumping step problem (abnormal jumping steps)