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:…
#Method1:动态规划##当有n个台阶时,可供选择的走法可以分两类:###1,先跨一阶再跨完剩下n-1阶:###2,先跨2阶再跨完剩下n-2阶.###所以n阶的不同走法的数目是n-1阶和n-2阶的走法数的和class Solution(object):    def climbStairs(self, n):        if n==1 or n==2 or n==0:            return n        steps=[1,1]        for i in xrang…
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:…
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:…
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…
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-…
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…
back function (return number) remember the structure class Solution { int res = 0; //List<List<Integer>> resList = new ArrayList<List<Integer>>(); public int combinationSum4(int[] nums, int target) { Arrays.sort(nums); return back(…
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…
leetcode 746. Min Cost Climbing Stairs(easy understanding dp solution) 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…
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:…
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…
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…
lc 746 Min Cost Climbing Stairs 746 Min Cost Climbing Stairs 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 t…
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…
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…
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…
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…
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…
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…
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. 题目分析:每次只能走1或2步,问n步的话有多少中走法???? 可以用动态规划和递归解…
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…
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:…
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; } };…
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…
[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…
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? 代码如下(方法一:超时): public class Solution { public int climbStairs(int n) { if(n==1||n==2|…