letecode [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

题目大意：

　　有n阶楼梯，可一次爬一阶或两阶，求共多少种爬法。

理  解 ：

　　n=0时，爬法=0；n=1时，爬法=1；n=2时，可爬2两次一阶、或一次两阶，爬法=2；

　　从n=3开始，设爬法为f，则f(n) = f(n-1) + f(n-2)；表示n阶阶梯的爬法等于当前爬一阶的种类+当前爬两阶的种类。

代 码 C++：　

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

运行结果：

　　执行用时 : 4 ms　　内存消耗 : 8.5 MB

标签：arr,int,top,step,steps,70,Climbing,climb,Stairs
来源： https://www.cnblogs.com/lpomeloz/p/10983623.html