1 概述 MySQL有两种方式可以实现ORDER BY: 1.通过索引扫描生成有序的结果 2.使用文件排序(filesort) 围绕着这两种排序方式,我们试着理解一下ORDER BY的执行过程以及回答一些常见的问题(下文仅讨论InnoDB存储引擎). 2 索引扫描排序和文件排序(filesort)简介 我们知道InnoDB存储引擎以B+树作为索引的底层实现,B+树的叶子节点存储着所有数据页而内部节点不存放数据信息,并且所有叶子节点形成一个(双向)链表. 举个例子,假设userinfo表的user

Given a nested list of integers, implement an iterator to flatten it. Each element is either an integer, or a list -- whose elements may also be integers or other lists. Example 1: Given the list [[1,1],2,[1,1]], By calling next repeatedly until hasN

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes. You should try to do it in place. The program should run in O(1) space complexi

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. 这道题是要求把有序链表转为二叉搜索树,和之前那道Convert Sorted Array to Binary Search Tree 将有序数组转为二叉搜索树思路完全一样,只不过是操作的数据类型有所差别,一个是数组,一个是链表.数组方便就方便在可以通过index直接访问任意一个元

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->2

02-线性结构1. 一元多项式求导 (25) 设计函数求一元多项式的导数.(注:xn(n为整数)的一阶导数为n*xn-1.) 输入格式:以指数递降方式输入多项式非零项系数和指数(绝对值均为不超过1000的整数).数字间以空格分隔. 输出格式:以与输入相同的格式输出导数多项式非零项的系数和指数.数字间以空格分隔,但结尾不能有多余空格.注意“零多项式”的指数和系数都是0,但是表示为“0 0”. 输入样例: 3 4 -5 2 6 1 -2 0 输出样例: 12 3 -10 1 6 0 最简单的方式是用

* Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ typedef struct ListNodep { int val; struct ListNodep *next; }ListNode; ListNode* addTwoNumbers(ListNode* I1, ListNode* I2) { ListNode *rootp = NULL;

在有些情况下,为了保持历史的一些状态,需要用拉链表来做,这样做目的在可以保留所有状态的情况下可以节省空间. 拉链表适用于以下几种情况吧 数据量有点大,表中某些字段有变化,但是呢变化的频率也不是很高,业务需求呢又需要统计这种变化状态,每天全量一份呢,有点不太现实, 不仅浪费了存储空间,有时可能业务统计也有点麻烦,这时,拉链表的作用就提现出来了,既节省空间,又满足了需求. 一般在数仓中通过增加begin_date,en_date来表示,如下例,后两列是start_date和end_date. 1 2

本文转载自:http://blog.csdn.net/coding__madman/article/details/51325646 链表简介: 链表是一种常用的数据结构,它通过指针将一系列数据节点连接成一条数据链.相对于数组,链表具有更好的动态性,建立链表时无需预先知道数据总量,可以随机分配空间,可以高效地在链表中的任意位置实时插入或者删除数据.链表的开销主要是访问的顺序性和组织链的空间损失. 内核链表的好主要体现为两点,1是可扩展性,2是封装.可扩展性肯定是必须的,内核一直都是在发展中的,所

题目来源:https://leetcode.com/problems/add-two-numbers/ You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a li

转载请注明出处为KlayGE游戏引擎,本文的永久链接为http://www.klayge.org/?p=2233 http://dogasshole.iteye.com/blog/1429665 http://www.gdcvault.com/ 2009年AMD在发布HD 5800的时候也发布了一个Order Independent Transparency(OIT)的demo,但只有介绍,没有多少可以参考的东西.GDC 2010上的OIT and GI using DX11 linked li

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 ->

题目: You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 -&

Linked List Sorting (链表)   A linked list consists of a series of structures, which are not necessarily adjacent in memory. We assume that each structure contains an integer key and a Next pointer to the next structure. Now given a linked list, you ar

Description 对于序列A,它的逆序对数定义为满足i<j,且Ai>Aj的数对(i,j)的个数.给1到n的一个排列,按照某种顺序依次删除m个元素,你的任务是在每次删除一个元素之前统计整个序列的逆序对数. Input 输入第一行包含两个整数n和m,即初始元素的个数和删除的元素个数.以下n行每行包含一个1到n之间的正整数,即初始排列.以下m行每行一个正整数,依次为每次删除的元素.   Output   输出包含m行,依次为删除每个元素之前,逆序对的个数. Sample Input 5 4 1

"Accordian" Patience You are to simulate the playing of games of ``Accordian'' patience, the rules for which are as follows: Deal cards one by one in a row from left to right, not overlapping. Whenever the card matches its immediate neighbour on

题目:给定单向链表的头指针和一个节点指针,定义一个函数在O(1)时间删除该节点. 链表节点与函数的定义如下. 通常我们删除某个节点都是从头开始遍历到需要删除节点的前一个节点. 然后使得该节点的next指向删除节点的next即可,这样看来删除一个节点 的复杂度为O(n)然而我们其实遍历的目的只是想获取想要删除节点的前一 个节点. 那么我们可以这样考虑: 我们把要删除节点下一个节点的值赋值到当前节点,然后将当前节点的下一个 节点删除即可. 比如: 一个链表3->2->5->7->9给定

这天要面试,提前把链表操作重新写了一遍.备份一下,以备不时之需. 希望有人能看到这篇代码,并指正. // File Name : list.h #include "stdafx.h" #include "stdio.h" #include <stdlib.h> #include "string.h" //creat list //creat node //insert //del //find //sort typedef int d

A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2521    Accepted Submission(s): 986 Problem Description Kiki likes traveling. One day she finds a magic lamp, unfortunately the genie

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 ->

1.Definition Linked list consists of a series of nodes. Each nodes contains the element and a pointer which points to the next node. The last node's next link points to NULL. Linked=data+pointer use the pointer to describe the logical relationship 2.