70. Climbing Stairs. Leetcode

Description:

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

Solution:

使用递归法

class Solution {
public:
    int climbStairs(int n) {
        return partOfStairs(n);
    }
    
    int partOfStairs(int n){
        if(n>2){
            return partOfStairs(n-1)+partOfStairs(n-2);
        } else if(n==2) {
            return 2;
        }else 
            return 1;
    }
};

难受

动态规划

类似斐波那契数列。
你只需要记住到达身后两个台阶的方法数，到达当前台阶的方法数就是，身后第一个台阶的向前一步一个台阶或者身后第二个台阶一步向前迈两台阶。
这样可知到达你当前的台阶方法数即为到达前面两个台阶方法数之和。
官方动图：

class Solution {
public:
    int climbStairs(int n) {
        int prev=1,cur=1,total=0; 
        for(int i = 2; i <= n; i++)
        {
            total = cur + prev; //Only last two element sum is required
            prev = cur;
            cur = total;
        }
        return cur;
    }
};