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:…

70. 爬楼梯 70. Climbing Stairs 题目描述 假设你正在爬楼梯.需要 n 阶你才能到达楼顶. 每次你可以爬 1 或 2 个台阶.你有多少种不同的方法可以爬到楼顶呢? 注意: 给定 n 是一个正整数. LeetCode70. Climbing Stairs 示例 1: 输入: 2 输出: 2 解释: 有两种方法可以爬到楼顶. 1 阶 + 1 阶 2 阶 示例 2: 输入: 3 输出: 3 解释: 有三种方法可以爬到楼顶. 1 阶 + 1 阶 + 1 阶 1 阶 + 2 阶 2…

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? 此题为典型的菲波那切数列问题: 当n=1时,有1种走法: 当n=2时,有2种走法; 当n=3时,有3种走法; 当n=4时,有5种…

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. 思路:到第n个台阶的迈法种数＝到第n-1个台…

题目描述 假设你正在爬楼梯.需要 n 阶你才能到达楼顶. 每次你可以爬 1 或 2 个台阶.你有多少种不同的方法可以爬到楼顶呢? 注意:给定 n 是一个正整数. 示例 1: 输入: 2 输出: 2 解释: 有两种方法可以爬到楼顶. 1. 1 阶 + 1 阶 2. 2 阶 示例 2: 输入: 3 输出: 3 解释: 有三种方法可以爬到楼顶. 1. 1 阶 + 1 阶 + 1 阶 2. 1 阶 + 2 阶 3. 2 阶 + 1 阶 思路 思路一: 递归 思路二: 用迭代的方法,用两个变量记录f(n-…

翻译 你正在爬一个楼梯. 它须要n步才干究竟顶部. 每次你能够爬1步或者2两步. 那么你有多少种不同的方法爬到顶部呢? 原文 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? 分析 动态规划基础题,首先设置3个变量用于…

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. Approach #1 Brute Force [Time Limit Ex…

https://leetcode.com/problems/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? 1)递归:  超时 class Solution { public…

lc 70 Climbing Stairs 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. D…

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:…

其实就是斐波那契数列 参考dp[n] = dp[n-1] +dp[n-2]; class Solution { public: int climbStairs(int n) { ; ; ; ; i < n; ++i){ f3 = f2 + f1; f1 = f2; f2 = f3; } return f3; } };…

1- 问题描述 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 can you climb to the top? 2- 思路分析 很自然的会想到递归,假设 n 级台阶有 T(n) 种不同走法.最后一步存在两种情况:剩下 1 级台阶,或者,剩下 2 级台阶. 第一种情…

题目描述: 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 can you climb to the top? 解题思路: 利用DP的方法,一个台阶的方法次数为1次,两个台阶的方法次数为2个.n个台阶的方法可以理解成上n-2个台阶,然后2步直接上最后一步:或者上n-1…

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? 思路:题目也比較简单.类似斐波那契. 代码例如以下: public class Solution { public int climb…

题目的意思是简化一个unix系统的路径.例如: path = "/home/", => "/home"path = "/a/./b/../../c/", => "/c" 我尝试用逐个字符判断的方法,一直提交测试,发现要修改甚多的边界.于是就参考了这位大神 思路其实不会那么复杂,C#里面的话直接可以用split就可以分割string,c++中好像要委婉实现,例如 getline(ss,now,'/') 在c++中ge…

假设你正在爬楼梯.需要 n 阶你才能到达楼顶. 每次你可以爬 1 或 2 个台阶.你有多少种不同的方法可以爬到楼顶呢? 注意:给定 n 是一个正整数. 示例 1: 输入: 2输出: 2解释: 有两种方法可以爬到楼顶.1. 1 阶 + 1 阶2. 2 阶 示例 2: 输入: 3输出: 3解释: 有三种方法可以爬到楼顶.1. 1 阶 + 1 阶 + 1 阶2. 1 阶 + 2 阶3. 2 阶 + 1 阶 设 $f[n]$ 表示跳上 $n$ 级台阶的方案数目,因此很容易得到 $f[n] = f[n-1…

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:…

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…

假设你正在爬楼梯.需要 n 阶你才能到达楼顶. 每次你可以爬 1 或 2 个台阶.你有多少种不同的方法可以爬到楼顶呢? 注意:给定 n 是一个正整数. 示例 1: 输入: 2 输出: 2 解释: 有两种方法可以爬到楼顶. 1. 1 阶 + 1 阶 2. 2 阶 示例 2: 输入: 3 输出: 3 解释: 有三种方法可以爬到楼顶. 1. 1 阶 + 1 阶 + 1 阶 2. 1 阶 + 2 阶 3. 2 阶 + 1 阶 思路 基本的动态规划问题,对于第n级台阶来说,有2种方法,1是到第n-1级,然…

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 Explanat…

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:…

[思路] a.因为两种跳法,1阶或者2阶,那么假定第一次跳的是一阶,那么剩下的是n-1个台阶,跳法是f(n-1); b.假定第一次跳的是2阶,那么剩下的是n-2个台阶,跳法是f(n-2) c.由a.b假设可以得出总跳法为: f(n) = f(n-1) + f(n-2) d.然后通过实际的情况可以得出:只有一阶的时候 f(1) = 1 ,只有两阶的时候可以有 f(2) = 2 e.可以发现最终得出的是一个斐波那契数列. 由于直接用递归会超时,于是用数组来存储每一个位置的走法数目.代码如下: cla…

题目描述 要爬N阶楼梯,每次你可以走一阶或者两阶,问到N阶有多少种走法 测试样例 Input: 2 Output: 2 Explanation: 到第二阶有2种走法 1. 1 步 + 1 步 2. 2 步 Input: 3 Output: 3 Explanation: 到第三阶有3种走法 1. 1 步 + 1 步 + 1 步 2. 1 步 + 2 步 3. 2 步 + 1 步 详细分析 在第0阶,可以选择走到第1阶或者第2阶,第1阶可以走第2阶或者第3阶,第二阶可以走第3阶或者第4阶....如此…

题目链接 题意 : 求斐波那契数列第n项 很简单一道题, 写它是因为想水一篇博客 勾起了我的回忆 首先, 求斐波那契数列, 一定 不 要 用 递归 ! 依稀记得当年校赛, 我在第一题交了20发超时, 就是因为用了递归, 递归时大量的出入栈操作必然比循环时间来得久 这题估摸着是每个测试样例就一个数, 记忆化的优势显示不出来, 但还是要认真看题 严格要求自己 记忆化搜索 vector<int> dp; int climbStairs(int n) { if (dp.size() <= 2)…

Leetcode 70 Climbing Stairs Easy https://leetcode.com/problems/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…

刷题备忘录,for bug-free leetcode 396. Rotate Function 题意: Given an array of integers A and let n to be its length. Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow: F(k…

[leetcode]51. N-QueensN皇后    Backtracking Hard [leetcode]52. N-Queens II N皇后 Backtracking Hard [leetcode]53. Maximum Subarray最大子数组和 Dynamic Programming Easy [leetcode]54. Spiral Matrix螺旋矩阵 Array Medium [leetcode]55. Jump Game青蛙跳(能否跳到终点) Greedy Medium…

Dynamic Programming 实际上是[LeetCode] 系统刷题4_Binary Tree & Divide and Conquer的基础上,加上记忆化的过程.就是说,如果这个题目实际上是类似于Divide and conquer或者说是DFS,但是在计算过程中有很多重复计算同样的过程的话,那么就可以用Dynamic prgramming/记忆化搜索来完成.基本就是利用空间来简化时间复杂度的过程. 可以/很有可能使用Dynamic programming的条件,满足之一即可. 1.…

老年退役选手的 LeetCode 休闲之旅 前言 不知不觉两年多的大学时光悄然流逝,浑浑噩噩的状态似乎从来没有离开过自己. 这两年刷题似乎一直是常态.在退役之后的现在,深感有些东西一段时间没有接触,很容易就变得陌生,遂萌生了刷 LeetCode 的想法,不知这一次能维持多久,谨以此记录来不时地警醒自己. -- 记于2018年寒假 LeetCode之旅 一开始的计划是先不考虑tag,把题目限制在 1~200 题,刷题的顺序是按照难度(Easy-Medium-Hard),题号(1~200). 结果一…

Note: 后面数字n表明刷的第n + 1遍, 如果题目有**, 表明有待总结 Conclusion questions: [LeetCode] questions conclustion_BFS, DFS LeetCode questions conclustion_Path in Tree [LeetCode] questions conlusion_InOrder, PreOrder, PostOrder traversal [LeetCode] questions for Dynamic…