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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).

Thinking: n-th order before a jump only two possibilities, either the first n-1, either the n-2, so f (n) = f (n-1) + f (n-2);

solution1:

1 class Solution {
2 public:
3     int jumpFloor(int number) {
4         if(number==1||number ==2)return number;
5       return jumpFloor(number-1)+jumpFloor(number-2);
6     }
7 };

Thoughts: recursion is very simple, find DP solution if that is positive to come and also no difference.