You are climbing a stair case. It takesnsteps to reach to the top.Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb...
转载
2014-11-14 09:12:00
81阅读
You are climbing a stair case. It takesnsteps to reach to the top.Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb...
转载
2013-11-12 14:24:00
124阅读
2评论
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 cl
原创
2022-01-12 10:37:02
18阅读
Climbing StairsYou are climbing a stair case. It takesnsteps to reach to the top.Each time you can either climb 1 or 2 steps. In how many distinct way...
原创
2021-08-07 14:11:17
122阅读
题目:题目链接题解:线性dp#include <bits/stdc++.h>using namespace std;long long a[20000
原创
2022-11-07 15:07:14
73阅读
原题链接: http://oj.leetcode.com/problems/climbing-stairs/ 这道题目是求跑楼梯的可行解法数量。每一步能够爬一格或者两个楼梯。能够发现。递推式是f(n)=f(n-1)+f(n-2)，也就是等于前一格的可行数量加上前两格的可行数量。熟悉的朋友可能发现了，
转载
2017-05-28 08:13:00
89阅读
2评论
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 cl
转载
2016-04-08 21:50:00
107阅读
2评论
LeetCode解题之Climbing Stairs 原题 一共同拥有n级楼梯，每次可以爬一级或两级。共同拥有多少种不同的爬法爬到顶端。 注意点： 无 样例： 输入: n = 6 输出: 13 解题思路 典型的动态规划题。递推表达式为 dp[i]=dp[i-1]+dp[i-2]。n为1时仅仅有一种方
原创
2021-12-31 09:37:21
21阅读
1.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?2.解决方案class Solution {public:
原创
2022-08-01 17:26:48
50阅读
Question 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?本题难度Easy。有3种算法分别是： 递归、迭代、数学
原创
2023-02-02 14:59:24
86阅读
You are climbing a stair case. It takesnsteps to reach to the top.Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb...
转载
2014-06-22 22:14:00
39阅读
2评论
这道题目是求跑楼梯的可行解法数量。每一步可以爬一格或者两个楼梯，可以发现，递推式是f(n)=f(n-1)+f(n-2)，也就是等于前一格的可行数量加上前两格的可行数量。熟悉的朋友可能发现了，这个递归式正是斐波那契数列的定义. 这里base case 是f(1)=1, f(2)=2. Fibonacc
转载
2014-05-20 07:10:00
99阅读
2评论
称号：给定一个int整数类型n，它代表了楼梯的阶。每一个可能的步骤时，楼梯，有可能采取两个步骤，求完成n楼梯有多少种不同的方法算法：递归是最简单的方法，但超时。递归转换的递推公式：f(n) = f(n-1)+f(n-2)public class Solution { public int cl...
转载
2015-06-26 11:19:00
70阅读
2评论
你正在爬楼梯。需要 n 步你才能到达顶部。每次你可以爬 1 或 2 个台阶。你有多少种不同的方式可以爬到楼顶呢？注意：给定 n 将是一个正整数。示例 1：输入： 2输出： 2说明： 有两种方法可以爬到顶端。1. 1 步 + 1 步2. 2 步示例 2：输入： 3输出： 3说明： 有三种方法可以爬到顶
转载
2018-04-03 12:32:00
48阅读
2评论
题目链接：https://leetcode.com/problems/climbing-stairs/题目：You are climbing a stair case. It takes n steps to reach to the top.Each
原创
2023-07-27 00:01:29
47阅读
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? 思路分析： 设 f (n) 表示爬 n 阶楼梯的不同方法数，
原创
2022-08-01 12:23:44
133阅读
You are climbing a stair case. It takesnsteps to reach to the top.Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb...
转载
2015-03-10 14:07:00
39阅读
2评论
题目描写叙述： 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 ca
转载
2017-05-25 14:01:00
53阅读
2评论
Climbing StairsTotal Accepted:55005Total Submissions:160176My Submissi
转载
2015-06-09 02:14:00
58阅读
2评论
https://oj.leetcode.com/problems/climbing-stairs/ http://blog.csdn.net/linhuanmars/article/details/23976963 public class Solution {
public int climbSta
原创
2015-01-04 11:06:58
349阅读
