[leetcode]70. Climbing Stairs爬楼梯

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.

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

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

 

题意：

有n阶台阶，可以每次爬1步或者2步， 爬到终点有多少种不同的方法

 

思路：

一维dp

用一个数组存下当前 k (k<=n)阶台阶有多少种不同的走法

发现规律

爬上n阶台阶 = 爬上n-1阶台阶再爬1阶 + 爬上n-2阶台阶再爬2阶

爬上n-1阶台阶 =  爬上n-2阶台阶再爬1阶  +  爬上n-3阶台阶再爬2阶

爬上n-2阶台阶 =  爬上n-3阶台阶再爬1阶  +  爬上n-4阶台阶再爬2阶

初始化，

dp[0] = 1

dp[1] = 1

转移方程，

dp[i] = dp[i-1] + dp[i-2] 

 

代码：

class Solution {
    public int climbStairs(int n) {
        int [] dp = new int[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];   
    }
}