[leetcode]Climbing Stairs

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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?

很简单的一维DP问题；

算法思路：当最后一步跨完台阶时候，有两种跨法：一步，或者两步。因此得到状态转移方程：f(n) = f(n - 1) + f(n - 2)；

【注意】直接用递归，会超时。

代码如下：

 1 public class Solution {
 2     public int climbStairs(int n) {
 3         if(n <= 0) return 0;
 4         if(n <= 2) return n;
 5         int[] hash = new int[n + 1];
 6         hash[1] = 1;
 7         hash[2] = 2;
 8         for(int i = 3; i <= n; i++){
 9             hash[i] = hash[i - 1] + hash[i - 2];
10         }
11         return hash[n];
12     }
13 }

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[leetcode]Climbing Stairs

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原文地址：http://www.cnblogs.com/huntfor/p/3883769.html