Brushing notes (6)-walking the stairs
Title description
A frog can jump up to 1 step at a time, or jump up to 2 steps. Find the total number of jumping methods for the frog to jump on an n-level step (different order to calculate different results).
Idea: Same as above. Thinking backwards, when the frog jumps to the nth step, there are two possibilities: one is from the n-1 step, and the other is from the n-2 step.
So the total probability = (n-1 case) + (n-2 case)
class Solution {
public:
int jumpFloor(int number) {
if(number==0)
return 0;
else if(number==1)
return 1;
else if(number==2)
return 2;
else
{
number--;
int a[2]={1,2};
while(number-1!=0)
{
int temp=a[1];
a[1]=a[0]+a[1];
a[0]=temp;
number--;
}
return a[1];
}
}
};