Description:

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.

Analisis:

這個問題的本質其實就是斐波那契數列， 爬到第n級樓梯的步數，是由f(n-1)與f(n-2) 所控制的。

則： f(1)=1, f(2)=2, f(n)=f(n-1)+f(n-2);

 1 int climbStairs(int n) {
 2    
 3      int zr=1;
 4      int st=2;
 5      int pri=zr, per=st;
 6      for(int i=3; i<=n;i++)
 7      {
 8           if (i & 0x001 == 1)  //odd
 9                pri=pri+per;
10           else
11                per=pri+per;
12      }
13      if(n & 0x001 ==1)
 
14        return pri;
15      else
16        return per;
17     }

[Leetcode 70]: Climbing Stairs