LeetCode 70. Climbing Stairs (Easy)

题目描述：

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：

Example 1:Input: 2Output:  2Explanation:  There are two ways to climb to the top.1. 1 step + 1 step2. 2 stepsExample 2:Input: 3Output:  3Explanation:  There are three ways to climb to the top.1. 1 step + 1 step + 1 step2. 1 step + 2 steps3. 2 steps + 1 step

题目大意：爬楼梯，一次可以爬一层或者两层，问爬到n层有多少种爬法

思路：终于有一道会做的dp了…..很感动，因为一眼看出来是个dp（状态的定义和转移很明显）。要达到n层，上一层一定在n-1层或者n-2层，以此推回去，就可以得到状态转移方程：dp[n] = dp[n - 1] + dp[n - 2]，第2层可以从第0层或者第1层到达，以此开始dp的迭代就可以了。

c++代码：

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