Leetcode 题目整理 climbing stairs】的更多相关文章

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. 对这种题目开始没有什么思路,借鉴博客http://blog.csdn.net/ken…
On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed). Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step w…
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? 题目意思: 上楼梯.假设要n步到达楼梯的顶部.每一次你只能上一个或两级台阶,问要到达顶部一共有多少种方法? 解题思路: 真是太巧了!!我…
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…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目大意 题目大意 解题方法 递归 记忆化搜索 动态规划 空间压缩DP 日期 [LeetCode] 题目地址:https://leetcode.com/problems/climbing-stairs/ Total Accepted: 106510 Total Submissions: 290041 Difficulty: Easy 题目大意 You are climbing a…
一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)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…
这是悦乐书的第159次更新,第161篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第18题(顺位题号是70).你正在爬楼梯,它需要n步才能达到顶峰.每次你可以爬1或2步,你可以通过多少不同的方式登顶?注意:给定n是一个正整数.例如: 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 输入:3 输出:3 说明:有三种方法可以爬到顶端 1.1步 + 1步 + 1步 2.1步 + 2步 3.2步 + 1步 输入:4 输出:5 说明:有5种方法…
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. 题解: 爬到第n层的方法要么是从第n-1层一步上来的,要不就是从n-2层2步…
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? Hide Tags: Dynamic Programming   Solution: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…
1.题目描述 2.问题分析 使用动态规划. 3.代码 int climbStairs(int n) { ){ return n; } ]; dp[] = ; dp[] = ; dp[] = ; ; i <= n ; i++){ dp[i] = dp[i-] + dp[i-]; } return dp[n]; }…
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? Subscribe to see which companies asked this question 简单的dp问题,代码如下: class Solution {…
9. Palindrome Number Determine whether an integer is a palindrome. Do this without extra space. click to show spoilers. Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the res…
1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + n…
一共有n个台阶,每次跳一个或者两个,有多少种走法,典型的Fibonacii问题 class Solution(object): def climbStairs(self, n): if n<0: return 0 if n<2: return 1 first,second = 1,1 for v in range(2,n+1): res = first+second first,second = return res 还有一种,每次可以跳任意阶,有2^(n-1)种跳法…
问题描述: On a staircase, the i-th step has some non-negative cost cost[i]assigned (0 indexed). Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the s…
368. Largest Divisible Subset 题意:找到所有元素都不同的数组中满足以下规则的最大子集,规则为:子集中的任意两个元素a和b,满足a%b=0或者b%a=0. 解答:利用动态规划法求解.先给数组排好序.定义dp[i]表示以nums[i]结尾的子集的大小,则dp[i+1]=dp[j]+1,if nums[i+1]%nums[j]==0, 0<=j<=i,或者dp[i+1]=1.时间复杂度为O(n2). 代码如下: public List<Integer> la…
Sqrt(x) Implement int sqrt(int x). Compute and return the square root of x. 注:这里的输入输出都是整数说明不会出现 sqrt(7)这种情况,思路一就是应用二分法进行查找.每次给出中间值,然后比对cur的平方与目标值的大小.需要先设定两个变量用来存放左右游标. 这里要考虑一下整型溢出的问题,另外,即使不能开出整数的也要近似给出整数部分,不能忽略. 代码如下: int Solution::mySqrt(int x) { //…
38. Count and Say The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, ... 1 is read off as "one 1" or 11. 11 is read off as "two 1s" or 21. 21 is read off as "one 2, then one 1&quo…
27. Remove Element Given an array and a value, remove all instances of that value in place and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. The order of elements can be changed.…
24. Swap Nodes in Pairs Given a linked list, swap every two adjacent nodes and return its head. For example,Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant space. You may not modify the…
20. Valid Parentheses Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]&quo…
14. Longest Common Prefix Write a function to find the longest common prefix string amongst an array of strings. 注:这题竟然连个示例都没有,说明特殊情况并不多,就是要找出所有字符串的最长公共前缀.他应该不会超过所有字符串中最短的那个,可以试着找出最短长度n,然后每个都读取和比较n个,并根据情况不断减小那个n.\ ) { return ""; } int n{int(strs…
今天的两道题关于基本数据类型的探讨,估计也是要考虑各种情况,要细致学习 7. Reverse Integer Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if…
1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + n…
目录 题目链接 注意点 解法 小结 题目链接 Climbing Stairs - LeetCode 注意点 注意边界条件 解法 解法一:这道题是一题非常经典的DP题(拥有非常明显的重叠子结构).爬到n阶台阶有两种方法:1. 从n-1阶爬上 2. 从n-2阶爬上.很容易得出递推式:f(n) = f(n-1)+f(n-2)于是可以得到下面这种最简单效率也最低的解法 -- 递归. class Solution { public: int climbStairs(int n) { if(n == 0 |…
On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed). Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step w…
1.题目 70. Climbing Stairs——Easy 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:…
一.题目说明 题目70. Climbing Stairs,爬台阶(楼梯),一次可以爬1.2个台阶,n层的台阶有几种爬法.难度是Easy! 二.我的解答 类似的题目做过,问题就变得非常简单.首先用递归方法计算: class Solution{ public: int climbStairs(int n){ if(n==1) return 1; if(n==2) return 2; return climbStairs(n-1) + climbStairs(n-2); } }; 非常不好意思,Tim…
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? 这个爬梯子问题最开始看的时候没搞懂是让干啥的,后来看了别人的分析后,才知道实际上跟斐波那契数列非常相似,假设梯子有n层,那么如何爬到第n层呢,因为每次只能怕1或2步,那…
July 28, 2015 Problem statement: 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? The problem is most popular question in the algorit…