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…

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…

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…

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…

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:…

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 |…

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…

目录 题目链接 注意点 解法 小结 题目链接 Min Cost Climbing Stairs - LeetCode 注意点 注意边界条件 解法 解法一:这道题也是一道dp题.dp[i]表示爬到第i层的最小cost,想要到达第i层只有两种可能性,一个是从第i-2层上直接跳上来,一个是从第i-1层上跳上来.所以可以得到dp[i] = min(dp[i- 2] + cost[i - 2], dp[i - 1] + cost[i - 1]).时间复杂度O(n). class Solution { pu…

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(…

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. 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…

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…

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…

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…

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…

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? 提示:Dinamic Programming,动态规划,这是个斐波那契数列问题. 方法一: //找规律,设n个台阶的方法数为f(n) //f(1) = 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? Example 1: Input: 2 Output: 2 Explanation: There are two ways to climb to the top. 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? Note: Given n will be a positive integer. Example 1: Input: 2 Output: 2 Explanat…

#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…

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…

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…

1. 记忆化搜索 - 自上向下的解决问题:使用vector来保存每次计算的结果,如果下次再碰到同样的需要计算的式子就不需要重复计算了. 2. 动态规划 - 自下向上的解决问题 解法一:自顶向下 解法二:自底向上 class Solution { private: vector<int> memo; int calcWays(int n){ ) ; //一个台阶都没有 ) ; //if(n==2) return 2; //有两种解决方法:一次迈一步,迈两次:一次迈两步 ) memo[n] = c…

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…

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; } };…