/* 1st method will lead to time limit *//* the time complexity is exponential sicne T(n) = T(n-1) + T(n-2) */ class Solution { /** * @param n: an integer * @return an integer f(n) */ public int fibonacci(int n) { // write your code here if (n == 1 ||…

描述 查找斐波纳契数列中第 N 个数. 所谓的斐波纳契数列是指: 前2个数是 0 和 1 . 第 i 个数是第 i-1 个数和第i-2 个数的和. 斐波纳契数列的前10个数字是: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34 ... public class Solution { /** * @param n: an integer * @return: an ineger f(n) */ public int fibonacci(int n) { // write your…

Find the Nth number in Fibonacci sequence. A Fibonacci sequence is defined as follow: The first two numbers are 0 and 1. The i th number is the sum of i-1 th number and i-2 th number. The first ten numbers in Fibonacci sequence is: 0, 1, 1, 2, 3, 5,…

Yet Another Source Code for LintCode Current Status : 232AC / 289ALL in Language C++, Up to date (2016-02-10) For more problems and solutions, you can see my LintCode repository. I'll keep updating for full summary and better solutions. See cnblogs t…

汇总贴 56. Two Sum[easy] 167. Add Two Numbers[easy] 53. Reverse Words in a String[easy] 82. Single Number[easy] 17. Subsets[medium] 18. Subsets II[medium] 219. Insert Node in Sorted Linked List[Naive] 366. Fibonacci[Naive] 452. Remove Linked List Elemen…

今天这篇博客就聊聊几种常见的查找算法,当然本篇博客只是涉及了部分查找算法,接下来的几篇博客中都将会介绍关于查找的相关内容.本篇博客主要介绍查找表的顺序查找.折半查找.插值查找以及Fibonacci查找.本篇博客会给出相应查找算法的示意图以及相关代码,并且给出相应的测试用例.当然本篇博客依然会使用面向对象语言Swift来实现相应的Demo,并且会在github上进行相关Demo的分享. 查找在生活中是比较常见的,本篇博客所涉及的这几种查找都是基于线性结构的查找.也就是说我们的查找表是一个线性表,我…

Difficulty: Easy Topic: Fibonacci seqs Write a function which returns the first X fibonacci numbers. ;; 首先实现一个求fibonacci数的函数 ;;最简单的实现,就是通过定义来实现递归函数,(如下的fibonacci-number),但是这样缺点很明显,首先这不算尾递归,代码里面有大量的重复计算.其次,jvm不支持尾调用优化,因此,即使是尾递归,当嵌套层侧过深时,也会出现stackoverf…

经典的Fibonacci数的问题 主要想展示一下迭代与递归,以及尾递归的三种写法,以及他们各自的时间性能. public class Fibonacci { /*迭代*/ public static int process_loop(int n) { if (n == 0 || n == 1) { return 1; } int a = 1, b = 1; int i = 1; while (i < n) { i++; int t = b; b = a + t; a = t; } return…

第一种:利用for循环 利用for循环时,不涉及到函数,但是这种方法对我种小小白来说比较好理解,一涉及到函数就比较抽象了... >>> fibs = [0,1] >>> for i in range(8): fibs.append(fibs[-2] + fibs[-1]) >>> fibs [0, 1, 1, 2, 3, 5, 8, 13, 21, 34] 或者说输入一个动态的长度: fibs = [0,1] num = input('How many…

Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…