Sword refers to the question of jumping steps for frogs

(1) A frog can jump up to 1 steps at a time, or jump up to 2 levels. Ask the frog to jump on an n-level step with a total number of hops. Recursive mode

   
  

    
   

   
public static int f(int n) { 
    
   

   
 //参数合法性验证 
    
   

   
 if (n < 1) { 
    
   

   
 System.out.println("参数必须大于1！"); 
    
   

   
 System.exit(-1); 
    
   

   
 } 
    
   

   
 if     ( n  ==    1    Span class= "pun" >| |   n  ==    2  )    return    1  ;     
    
   

   
 else return f(n - 1) + f(n - 2); 
    
   

   
 } 
   
  

Non-recursive mode

  
 

   
  

  
public static int fx(int n) { 
   
  

  
//参数合法性验证 
   
  

  
if (n < 1) { 
   
  

  
System.out.println("参数必须大于1！"); 
   
  

  
System.exit(-1); 
   
  

  
} 
   
  

  
//n为1或2时候直接返回值 
   
  

  
if (n< 2) return 1; 
   
  

  
//n>2时候循环求值 
   
  

  
int res = 0; 
   
  

  
int a = 1; 
   
  

  
int b = 1; 
   
  

  
for (int i = 2; i <= n; i++) { 
   
  

  
res = a + b; 
   
  

  
a = b; 
   
  

  
b = res; 
   
  

  
} 
   
  

  
return res; 
   
  

  
} 
  
 

(2) A frog can jump up to 1 steps at a time, can also jump on 2 levels ... It can also jump on the N-level, at which point the frog jumps up the level of the N-step total number of hops?

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:

  
 

   
  

  
 if(number<2)return 1;
   
  

  
 //n>2时候循环求值
   
  

  
 int res = 0;
   
  

  
 int a = 1;
   
  

  
 for (int i = 2; i <= number; i++) {
   
  

  
 res = 2*a;
   
  

  
 a= res;
   
  

  
 }
   
  

  
 return res; 
  
 

From for notes (Wiz)

Sword refers to the question of jumping steps for frogs