You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Note: Given n will be a positive integer.
Example 1:
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps.
Example 2:
Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step
一个数只能由1和叠加起来，例如5可以由这样叠加起来：
结果可按普通的递归公式得到，但根据规律发现结果与斐波那契数列结果相同，所以使用此方法也可得到结果。
代码1：使用普通方法
int climbStairs(int n) {
return climb_Stairs(0,n);
}
int climb_Stairs(int i,int n)
{
if(i>n)
return 0;
if(i==n)
return 1;
return climb_Stairs(i+1,n)+climb_Stairs(i+2,n);
}
代码2：使用斐波那契数列
int climbStairs1(int n) {
if(n==1)
return1;
int first=1,second=2,third;
for(int i=3;i<=n;i++)
{
third=first+second;
first=second;
second=third;
}
return second;
}