资源列表
1006194
- 使用的依赖包都在压缩包的lib目录下,如果只是需要依赖包()
实验七
- 升级文本编辑器： 要求：支持菜单选取文本和背景颜色，支持弹出控制面板设置文本字体大小和颜色(Update text editor: Requirements: support menu, select text and background color, support pop-up control panel, set text, font size and color)
PrintZigZag
- 锯齿形的转换 字符串“paypalishiring”写的是在一个给定的行数，这样一个曲折的模式：（你可能想在一个固定的字体更加易读显示模式）(ZigZag Conversion The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed fon
渐变柱状图代码
- matlab编程，生成渐变色的柱状图，可增强文章的可读性(bars in gradient color)
LongestSubstr
- 从一个字符串中找到一个连续子串，该子串中任何两个字符不能相同，求子串的最大长度并输出一条最长不重复子串。(From a string to find a continuous substring, any two characters cannot be the same substring, the maximum length of her series and output a maximum repeat substring.)
findMedianSortedArrays
- 给定两个数组大小分别为m和n，排好了序，可能是降序也可能是升序，求两个数组所有数字的中位数，要求算法复杂度为O(m+n)。这里的中位数是如下定义的：如果总个数为偶数那么就取第n/2和n/2+1个数的平均数，例如： 两个数组分别为：[1,2] 和[1,2]那么中位数就应该是1,1,2,2的中位数，也就是：1.5(Given two array sizes, respectively, m and N, arranged in order, may be descending, or may be
Power
- 解题方法主要是利用递归和二分，具体就是把x的n次方划分成两个x的n/2次方相乘，然后递归求解子问题，结束条件是n为0返回1。因为是对n进行二分，所以算法复杂度是O(logn)。(Problem solving method is mainly the use of recursive and two points, specifically the x n side is divided into two X n/2 times multiplication, and then recursiv
Arduino-Temperature-Control-Library-master
- This technique is intended as a work around when you have been left with a password protected PLC and the original installer has gone bust
FlowD
- 绘制流程图，然后可根据流程图计算拓扑图顺序，并按照顺序执行(According to the calculation of topological order, and in accordance with the order of execution)
sednkv
- 分行原理编写的程序,用JAVA编写,希望对大家有所帮助,()
leng-V8.0
- It uses a pulse of consumer law, Based on SVPWM three-level inverter matlab simulation, Really is a good program.
50389950
- 利用java语言编制的一些排序源代码,如气泡排序法等等,()