算法70. Climbing Stairs

70. 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?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

爬楼梯。有n阶楼梯。
每次只能爬一阶或者两阶。问，有多少种方式爬到顶？
注意：
n 只会是正整数。

解：
又一个典型的使用动态规划的解决的问题。要爬到 n 阶楼梯，无非两种爬法：要么从 n-1 爬上来，要么从 n-2 爬上来，以此为原型设计动态规划模型。以下为代码：

public int climbStairs(int n) {
    if (n == 1) {// 为1时，只有一种爬法
        return 1;
    }
    int[] dp = new int[n + 1];
    dp[1] = 1;
    dp[2] = 2;
    for (int i = 3; i <= n; i++) {
        dp[i] = dp[i - 1] + dp[i - 2];// 为n时，爬法为 n-1 与 n-2 爬法的和
    }
    return dp[n];
}