【Leetcode 70】关关的刷题日记68 – Leetcode 70 Climbing Stairs

关关的刷题日记68 – Leetcode 70 Climbing Stairs

题目

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

题目的意思是爬楼梯，楼梯一共有n个台阶。每次可以爬一个台阶或者两个台阶，问爬到楼顶一共有多少种方式。

思路

思路：动态规划的题目：假设爬到第i个楼梯一共有re[i]种方式，则状态转移方程re[i]= re[i-1]+ re[i-2]，即爬到第i个楼梯可以是从第i-1个楼梯，爬一步上来的，也可以是从第i-2个楼梯爬两步上来。

class Solution {public:
    int climbStairs(int n) {
        if(n<=2)
            return n;
        int temp1=1, temp2=2, re;
        for(int i=3; i<=n; i++)
        {
            re=temp1+temp2;
            temp1=temp2;
            temp2=re;
        }
        return re;
    }};

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