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搜索关键字:minimum inversion nu    ( 5304个结果
MSSQL 获取数据库字段类型
1 SELECT 2 col.name AS 列名, 3 typ.name as 数据类型, 4 col.max_length AS 占用字节数, 5 col.precision AS 数字长度, 6 col.scale AS 小数位数, 7 col.is_nu...
分类:数据库   时间:2014-06-27 16:39:34    阅读次数:291
LeetCode:Triangle
Given a triangle, find the minimum path sum from top to bottom. Each step you maymove to adjacent numbers on the row below.For example, given the fol....
分类:其他好文   时间:2014-06-27 16:29:58    阅读次数:188
mysql 索引 使用注意细节
在查询时,如果使用到LIKE关键字,就要注意有没有使用到索引;没有使用索引前:mysql>explainselect*fromemployeeswherefirst_name=‘Georgi‘\G;***************************1.row***************************id:1select_type:SIMPLEtable:employeestype:ALLpossible_keys:NU..
分类:数据库   时间:2014-06-27 06:10:07    阅读次数:272
1.7 逆序数与归并排序[inversion pairs by merge sort]
inversion pairs by merge sort
分类:其他好文   时间:2014-06-26 23:34:49    阅读次数:294
js加强小结
一)回顾JavaScript基础 (1)函数的定义方式 *>>正常方式 function add(num1,num2){...} >>构造器方式 var add = new Function("num1","num2","return num1+nu...
分类:Web程序   时间:2014-06-26 22:15:17    阅读次数:225
LeetCode: Palindrome Partitioning II [132]
【题目】 Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s = "aab", Return 1 since the palindrome partitioning ["aa","b"] could b...
分类:其他好文   时间:2014-06-26 07:48:00    阅读次数:259
Qt5 Cmake
project(my)cmake_minimum_required(VERSION 2.8.9)set (CMAKE_PREFIX_PATH "C:\\Qt\\Qt5.3.0\\5.3\\msvc2010_opengl")set(CMAKE_INCLUDE_CURRENT_DIR ON)find_p...
分类:其他好文   时间:2014-06-25 16:59:54    阅读次数:221
Codeforces Round #253 (Div. 2), problem: (B)【字符串匹配】
简易字符串匹配,题意不难 1 #include 2 #include 3 #include 4 #include 5 #include 6 using namespace std; 7 8 int main(){ 9 int i, j, k, t, n;10 int nu...
分类:其他好文   时间:2014-06-25 12:07:42    阅读次数:145
leetcode--Edit Distance
Given two wordsword1andword2, find the minimum number of steps required to convertword1toword2. (each operation is counted as 1 step.)You have the fol...
分类:其他好文   时间:2014-06-25 11:14:15    阅读次数:218
Raphael.js API之Raphael.pathIntersection(),aphael.pathToRelative(),Set.clear(),Set.exclude(element)
/*API-141*/Raphael.pathIntersection(path1, path2)获取两条线的交点参数列表:path1 字符串类型 路径的字符串表达形式path2 字符串类型 路径的字符串表达形式返回值:交点集合,格式如下:[{ x: //number类型 点的x坐标 y: //nu...
分类:Windows程序   时间:2014-06-24 15:17:00    阅读次数:233
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