算法题之Climbing Stairs（leetcode 70）

【算法题之Climbing Stairs（leetcode 70）】的更多相关文章

算法题之Climbing Stairs（leetcode 70）

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…

Min Cost Climbing Stairs - LeetCode

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…

Climbing Stairs - LeetCode

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 |…

Cllimbing Stairs [LeetCode 70]

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…

[面试算法题]比较二叉树异同-leetcode学习之旅（5）

问题描述 Given two binary trees, write a function to check if they are equal or not. Two binary trees are considered equal if they are structurally identical and the nodes have the same value. 和Balanced Binary Tree那题很像.定义T(p)为以p为根的树,T(p) = T(q)需要符合两个条件:…

climbing stairs leetcode java

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;…