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:…
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…
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…
网址:https://leetcode.com/problems/climbing-stairs/ 其实就是斐波那契数列,没什么好说的. 注意使用3个变量,而不是数组,可以节约空间. class Solution { public: int climbStairs(int n) { ) return n; , c = ; ;i<=n;i++) { a = b + c; c = b; b = a; } return a; } };…
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-…
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…
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…
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…
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:…
题目描写叙述: 初阶:有n层的台阶,一開始你站在第0层,每次能够爬两层或者一层. 请问爬到第n层有多少种不同的方法? 进阶:假设每次能够爬两层.和倒退一层,同一个位置不能反复走,请问爬到第n层有多少种不同的方法? 解题思路: 初阶:一维动态规划.爬楼梯数目事实上是一个斐波拉契数列. 假定f[i] 表示是爬到第i层的方法,那么f[i] = f[i-1] + f[i-2] //第i层的方法数目等于第i-1层数目加上第i-2层数目. 初值:f[0] = 1, f[1] = 1. //最開始没有爬和第一…
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…
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? 题意:有n阶楼梯,每次可以走一步或者两步,总共有多少种方法. 思路:动态规划.维护一个一维数组dp[n+1],dp[0]为n=0时的情况,dp[ i ]为到达第i阶台阶…
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? 当n=1时,有1种方法,即直接走1步 当n=2时,有2方法:连续走2步,或直接走两步 对于n,设f(n)为总方法,则 f(n) = f(n-1)+f(n-2) ps:f…
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…
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; } };…
Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the nodes in each of the two partitions. For example, Given 1->4->3->…
Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue. Here, we will use the integers 0, 1, and 2 to represent the color red, white, and bl…
3Sum  Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. Note: Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c) The…
Insert Interval  Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary). You may assume that the intervals were initially sorted according to their start times. Example 1: Given intervals [1,3],[6,9],…
Rotate List  Given a list, rotate the list to the right by k places, where k is non-negative. For example: Given 1->2->3->4->5->NULL and k = 2, return 4->5->1->2->3->NULL. 思路:题目非常清晰.思路是先得到链表长度.再从头開始直到特定点,開始变换连接就可以. 代码例如以下: /*…
Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100". 思路:二进制加法,比較简单.代码例如以下: public class Solution { public String addBinary(String a, String b) { int len = Math.max(a.l…
Spiral Matrix Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. For example, Given the following matrix: [  [ 1, 2, 3 ],  [ 4, 5, 6 ],  [ 7, 8, 9 ] ] You should return [1,2,3,6,9,8,7,4,5]. 思路:螺旋数…
Plus One Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. 思路:给定一个数组,表示一个数.然后返回+1的值.主要就是进位的问题.代码例如以下: public class Solution {…
Remove Element Given an array and a value, remove all instances of that value in place and return the new length. The order of elements can be changed. It doesn't matter what you leave beyond the new length. 思路:此题和26题一脉相承,算法上不难,详细如代码所看到的: public clas…