leetcode70—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. Example 1: Input: 2 Output: 2 Explanation:…
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? (Easy) 分析: 经典的爬楼梯问题,对于第n阶可以从n - 1阶跳一步上来,也可以从n - 2阶跳两步上来,设dp[n]为到第n阶的方案数, 则dp[n]…
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…
题目 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 ste…
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…
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? Have you met this question in a real interview? Yes Example Given an example n=3 , 1…
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…
Search a 2D Matrix Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted from left to right. The first integer of each row is greater than the last integer…
Climbing Stairs 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? 算法描述 (1)这道题其实是一道fibon…
原题 Climbing Stairs 求斐波那契数列的第N项,开始想用通项公式求解,其实一个O(n)就搞定了. class Solution { public: int climbStairs(int n) { ) ; ; ; ; i<n; ++i) { int t = n2; n2 = n1 + n2; n1 = t; } return n2; } }; 还是觉得没有ACM难度,继续做吧.…
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…
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…
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…
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…
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-…
[抄题]: 假设你正在爬楼梯,需要n步你才能到达顶部.但每次你只能爬一步或者两步,你能有多少种不同的方法爬到楼顶部? [思维问题]: 不知道一步.两步怎么加.还是用iteration迭代.此题公式可被称为斐波那契数列. 不知道和坐标型有什么关系:列j = 1即可 [一句话思路]: 生兔子问题:把大数赋给小数 [输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): [画图]: [一刷]: [二刷]: [三刷]: [四刷]: [五刷]: [五分钟肉眼debu…
Climbing Stairs https://oj.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? 这题比较简单,可以使用动态规划来求解…
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…
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 |…
Climbing Stairs 本题收获: 1.斐波那契函数f(n) = f(n-1) + f(n -2) 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? 有n阶台阶,每次可以走2步或者1步,问有多少种方法到…
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(…
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…
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…
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…
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:…
746. 使用最小花费爬楼梯 746. Min Cost Climbing Stairs 题目描述 数组的每个索引做为一个阶梯,第 i 个阶梯对应着一个非负数的体力花费值 cost[i].(索引从 0 开始) 每当你爬上一个阶梯你都要花费对应的体力花费值,然后你可以选择继续爬一个阶梯或者爬两个阶梯. 您需要找到达到楼层顶部的最低花费.在开始时,你可以选择从索引为 0 或 1 的元素作为初始阶梯. 每日一算法2019/5/14Day 11LeetCode746. Min Cost Climbing…
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: Input:…
problem 746. Min Cost Climbing Stairs 题意: solution1:动态规划: 定义一个一维的dp数组,其中dp[i]表示爬到第i层的最小cost,然后来想dp[i]如何推导.思考一下如何才能到第i层呢?是不是只有两种可能性,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.不会再有别的方法,所以dp[i]只和前两层有关系,所以可以写做如下: dp[i] = min(dp[i- 2] + cost[i - 2], dp[i - 1] + cost[i…