【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?
分析：
简单的动态规划问题。求整数n通过不断减1或者2有几种方法到0.
状态转移方程为：ways[x]=ways[x-1]+ways[x-2].
算法的复杂度为O（n）。
代码：

class Solution {public:    int climbStairs(int n) {        int *ways = new int[n+1];         ways[0] = 0;        ways[1] = 1;        ways[2] = 2;        for(int i = 3;i<=n;i++)        {            ways[i] = ways[i-1]+ways[i-2];        }        return ways[n];    }};