LeetCode -- 70. Climbing Stairs

题目：

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.

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思路：
这道题很简单，最简单的动态规划，也是一个斐波那契数列。
定义d[i]为i阶楼梯的方法数，那么到达第i阶楼梯只有两种方法，从i-1阶楼梯上一个台阶，从i-2个台阶上两个楼梯。
初始状态：

d[0]=d[1]=1;

状态转移方程：

d[i]=d[i−1]+d[i−2],i>=2;

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C++代码:

class Solution {public:    int climbStairs(int n) {        int d[100];        d[0]=1, d[1]=1;        for(int i=2;i<=n;i++)        {            d[i] = d[i-1]+d[i-2];        }        return d[n];    }};