Prove safety offer8: frog jump stairs

1. Title Description

　　A frog can jump on a Class 1 level, you can also hop on level 2. The frog jumped seeking a total of n grade level how many jumps (the order of different calculation different results).

2. The ideas and methods

　　Each jump frog only two options: Level 1 jump reaches the n-th stage ladder step, this time in the first-stage small frogs step n-1; jump or stage 2 reaches the n-th stage ladder step, this time a small frog It is n-2-step . Thus, n jump-step process always depends on the total number n-1 pre-step jump method F (n-1) n-2 and the front-step jump method Total f (n-2). Because only two possibilities, therefore, f (n) = f ( n-1) + f (n-2);

　　Recurrence formula f (n) = f (n-1) + f (n-2): very familiar, is the Fibonacci Summation verify, there is one kind of grade 1 frog jumps, there are two kinds of level 2 jump method, there are three three jump method, there are five four jumps.

3. C ++ core code

3.1 Recursive (inefficient, time complexity O (n- ² ))

 1 class Solution {
 2 public:
 3 int jumpFloor(int number) {
 4         if(number==1)
 5             return 1;
 6         else if(number==2)
 7             return 2;
 8         else{
 9             return jumpFloor(number-1)+jumpFloor(number-2);
10         }
11     }
12 };

View Code

3.2 nonrecursive (time complexity of O (n))

 1 class Solution {
 2 public:
 3     int jumpFloor(int number) {
 4         if (number < 0)
 5         {
 6             return 0;
 7         }
 8         if (number == 0 || number == 1 || number == 2)
 9         {
10             return number;
11         }
12         int f1 = 1;
13         int f2 = 2;
14         int result = 0;
15         for (int i = 3; i <= number; i++)
16         {
17             result = f1 + f2;
18             f1 = f2;
19             f2 = result;
20         }
21         return result;
22     }
23 };

View Code

Reference material

https://blog.csdn.net/qq_33022911/article/details/83536283