2016.12.4, 366 http://www.lintcode.com/en/problem/fibonacci/ 一刷使用递归算法,超时.二刷使用九章算术的算法,就是滚动指针的思路,以前写python的时候也玩过,但是给忘了,这次又用c++拾起来了.lint有bug,不能用,很烦. class Solution { public: /** * @param n: an integer * @return an integer f(n) */ int fibonacci(int n) {…

刷题备忘录,for bug-free leetcode 396. Rotate Function 题意: Given an array of integers A and let n to be its length. Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow: F(k…

463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…

刷题备忘录,for bug-free 招行面试题--求无序数组最长连续序列的长度,这里连续指的是值连续--间隔为1,并不是数值的位置连续 问题: 给出一个未排序的整数数组,找出最长的连续元素序列的长度. 如: 给出[100, 4, 200, 1, 3, 2], 最长的连续元素序列是[1, 2, 3, 4].返回它的长度:4. 你的算法必须有O(n)的时间复杂度 . 解法: 初始思路 要找连续的元素,第一反应一般是先把数组排序.但悲剧的是题目中明确要求了O(n)的时间复杂度,要做一次排序,是不能达…

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…

--------------------------------------------------------------- 本文使用方法:所有题目,只需要把标题输入lintcode就能找到.主要是简单的剖析思路以及不能bug-free的具体细节原因. ---------------------------------------------------------------- ------------------------------------------- 第九周:图和搜索. ---…

Implement a trie with insert, search, and startsWith methods. Have you met this question in a real interview?     Example Note You may assume that all inputs are consist of lowercase letters a-z. LeetCode上的原题,请参见我之前的博客Implement Trie (Prefix Tree) 实现字…

原题链接在这里:http://www.lintcode.com/en/problem/subtree/ You have two every large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree ofT1. Have you met this question in a real intervi…

Given an array of integers and a number k, the majority number is the number that occurs more than 1/k of the size of the array. Find it. Have you met this question in a real interview? Yes Example Given [3,1,2,3,2,3,3,4,4,4] and k=3, return 3. Note…

A peak element is an element that is greater than its neighbors. Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the pea…

http://www.lintcode.com/zh-cn/problem/topological-sorting/# 给定一个有向图,图节点的拓扑排序被定义为: 对于每条有向边A--> B,则A必须排在B之前 拓扑排序的第一个节点可以是任何在图中没有其他节点指向它的节点 找到给定图的任一拓扑排序 solution Topological Sorting Topological sorting for Directed Acyclic Graph (DAG) is a linear orderi…

Longest Common Subsequence 原题链接:http://lintcode.com/zh-cn/problem/longest-common-subsequence/ Given two strings, find the longest comment subsequence (LCS). Your code should return the length of LCS. 样例For "ABCD" and "EDCA", the LCS is…

Majority Number II 原题链接: http://lintcode.com/en/problem/majority-number-ii/# Given an array of integers, the majority number is the number that occurs more than 1/3 of the size of the array. Find it. Note There is only one majority number in the arra…

Kth Largest Element Find K-th largest element in an array. Note You can swap elements in the array Example In array [9,3,2,4,8], the 3th largest element is 4 Challenge O(n) time, O(1) space 原题链接: http://www.lintcode.com/en/problem/kth-largest-element…

First Bad Version http://lintcode.com/en/problem/first-bad-version The code base version is an integer and start from 1 to n. One day, someone commit a bad version in the code case, so it caused itself and the following versions are all failed in the…

studio配置CodeStyle可以很好的帮助我们检测代码规范性,保持大家的代码统一,来看看怎么配置和使用吧 代码规范,自己公司的一套 代码规范 一.      简介 A.    目的 本文提供一整套编写高效可靠的 Java代码的标准.约定和指南.它们以安全可靠的软件工程原则为基础,使代码易于理解.维护和增强.而且,通过遵循这些程序设计标准,你作为一个 Java软件开发者的生产效率会有显著提高.经验证明,若从一开始就花时间编写高质量的代码,则在软件开发阶段,对代码的修改要容易很多.最后,遵循一…

Alex and Lee play a game with piles of stones.  There are an even number of piles arranged in a row, and each pile has a positive integer number of stones piles[i]. The objective of the game is to end with the most stones.  The total number of stones…

Functionals “To become significantly more reliable, code must become more transparent. In particular, nested conditions and loops must be viewed with great suspicion. Complicated control flows confuse programmers. Messy code often hides bugs.” — Bjar…

------------------------------------------------------------第七周:Follow up question 1,寻找峰值 寻找峰值 描述 笔记 数据 评测 你给出一个整数数组(size为n),其具有以下特点: 相邻位置的数字是不同的 A[0] < A[1] 并且 A[n - 2] > A[n - 1] 假定P是峰值的位置则满足A[P] > A[P-1]且A[P] > A[P+1],返回数组中任意一个峰值的位置. 注意事项 数…

[题目描述] 根据前序遍历和中序遍历树构造二叉树. 在线评测地址: https://www.jiuzhang.com/solution/construct-binary-tree-from-preorder-and-inorder-traversal/?utm_source=sc-bky-zq [样例] 样例 1: 输入:[],[] 输出:{} 解释: 二叉树为空 样例 2: 输入:[,,],[,,] 输出:{,,} 解释: 二叉树如下 / \ [题解] Given preorder and i…

三星Note 7发售两个月即成为全球噩梦,从首炸到传言停产仅仅47天.所谓"屋漏偏逢连天雨",相比华为.小米等品牌对其全球市场的挤压.侵蚀,Galaxy Note 7爆炸事件这场连天雨算是把三星淋了个措手不及,致其声誉扫地.危机重重. 表面看来,此事件源于电池自燃,深究起来可以追溯到管理层变动带来的组织管理问题,再深究,业务流程管理系统可能是最基础和最关键的因素. 高层变动,赶超苹果埋祸端 2014年5月,三星老掌柜李健熙重病入院,儿子李在镕与元老崔志成(G.S.Choi)共掌时局.同…

Mem pro 是一个主要集成内存泄露检测的工具,其具有自身的源码和GUI,在GUI中利用"Launch" button进行加载自己待检测的application,目前支持的平台为Windows,Unix, Linux, OSX, IOS, GCC:但是按照官网的说法,其虽然只能运行到WIN上,但是根据TCP协议传输dump的方式也可以和其他平台的app进行连接: 关于内存泄露,按照官方文档中的说法,其检测内存泄露的算法主要是两种,一种是在抓取dump时候未被引用的变量会被认定为泄露,…

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 整体模块module分析: 打开Strate…

环境 Chrome jmeter3.1 fiddler4 win7 32位 Linux CentOs6.4 bugfree3.0.1 链接:http://pan.baidu.com/s/1gfHpbpD 密码:wpxj  jmeter3.1软件包 链接:http://pan.baidu.com/s/1miHeNfa 密码:yqlp  fiddler4软件包 如果链接失效,可以自己网上寻找资源,或者加我百度云:天涯咫尺HK 一.打开jmeter.添加线程组.录制控制器.HTTPCookie管理器.…


Given
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 note
 string 
and 
another 
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 all 
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 write 
a 
function 
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will 
return 
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the 
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 note 
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the 
magazines ; 
otherwise, 
it 
wil…

-------------------------------------------- AC代码: /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ pub…

----------------------------------- Moore's voting algorithm算法:从一个集合中找出出现次数半数以上的元素,每次从集合中去掉一对不同的数,当剩下一个元素的时候(事实上只要满足一个元素出现过半就一定会剩下一个元素的)这个元素就是我们要找的数了. AC代码: public class Solution { /** * @param nums: a list of integers * @return: find a majority numb…

----------------------------------- 最开始的想法是先计算出链表的长度length,然后再从头走 length-n 步即是需要的位置了. AC代码: /** * Definition for ListNode. * public class ListNode { * int val; * ListNode next; * ListNode(int val) { * this.val = val; * this.next = null; * } * } */ pu…

------------------------ 因为字符究竟是什么样的无法确定(比如编码之类的),恐怕是没办法假设使用多大空间(位.数组)来标记出现次数的,集合应该可以但感觉会严重拖慢速度... 还是只做出了O(n^2)... 勉强AC代码: public class Solution { /** * @param str: a string * @return: a boolean */ public boolean isUnique(String s) { for(int i=0;i<s.…

-------------------- 递归那么好为什么不用递归啊...我才不会被你骗...(其实是因为用惯了递归啰嗦的循环反倒不会写了...o(╯□╰)o) AC代码: /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left…