【leetcode】Climbing Stairs

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?

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题解，变相的斐波那契数列。因为最后一步到达n的时候有两种选择，走1步走到台阶n和走2步走到台阶n。如果走一步走到台阶n，那么之前的n-1级台阶有f(n-1)种走法；如果走2步走到台阶n，那么之前的n-2级台阶有f(n-2)种走法，所以一共有f(n-1)+f(n-2)步走法。

代码如下：

 1 class Solution {

 2 public:

 3     int climbStairs(int n) {

 4         vector<int> num_method(n+1);

 5         num_method[0] = num_method[1] = 1;

 6         for(int i = 2;i <= n;i ++)

 7             num_method[i] = num_method[i-1]+num_method[i-2];

 8         return num_method[n];

 9     }

10 };

 Java代码:不需要数组，只需要保存前面两级台阶的走法数即可。

 1 public class Solution {

 2     public int climbStairs(int n) {

 3         if(n == 1)

 4             return 1;

 5         if(n == 2)

 6             return 2;

 7         int first = 1;

 8         int second = 2;

 9         int answer = 0;

10         for(int i = 3;i <= n;i++)

11         {

12             answer = first + second;

13             first = second;

14             second = answer;

15         }

16         return answer;

17     }

18 }