title: UP board 漫谈(1)--从Atom到UP Board date: 2016-12-26 12:33:03 tags: UP board categories: 开发板 permalink: upboard_ramble --- 前言 原创文章,转载引用务必注明链接.如有疏漏,欢迎指正. 图文部分引用自CNXSoft ​ 每块开发板都有其设计理念,也是其特色所在.有做工优良.接口丰富的多媒体全能开发板Lemaker Guitar ,有专注物联网开发的Intel Edison…

前言 原创文章,转载引用务必注明链接.如有疏漏,欢迎指正. 图文部分引用自CNXSoft ​ 每块开发板都有其设计理念,也是其特色所在.有做工优良.接口丰富的多媒体全能开发板Lemaker Guitar ,有专注物联网开发的Intel Edison ,有支持专业应用的Beaglebone Black ,还有可以学习编程的树莓派(需要拓展板支持才能完成更复杂的任务). ​ 那我们再来看看UP board.它于2015年10月16日在Kickstarter上众筹,从众筹页面上可以看到主要针对的是树莓…

Given an 2D board, count how many different battleships are in it. The battleships are represented with 'X's, empty slots are represented with '.'s. You may assume the following rules: You receive a valid board, made of only battleships or empty slot…

https://leetcode.com/problems/battleships-in-a-board/ 给定一个N×N的棋盘,有任意数量的1×N或N×1大小的"船",注意船船之间是不相邻的,要求统计有多少船 Example: X..X ...X ...X In the above board there are 2 battleships. 解题思路 Given an 2D board, count how many different battleships are in it.…

The follow-up question is fun: "Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?" When we meet an 'X', we need to check if it is vertical or horizontal: they will never happen at the same ti…

A Board Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 551   Accepted: 373 Description Dao was a simple two-player board game designed by Jeff Pickering and Ben van Buskirk at 1999. A variation of it, called S-Dao, is a one-player…

Description Dao was a simple two-player board game designed by Jeff Pickering and Ben van Buskirk at . A variation of it, called S-Dao, * square with cells. There are black stones and white stones placed on the game board randomly in the beginning. T…

Given an 2D board, count how many battleships are in it. The battleships are represented with 'X's, empty slots are represented with '.'s. You may assume the following rules: You receive a valid board, made of only battleships or empty slots. Battles…

［抄题］: Given an 2D board, count how many battleships are in it. The battleships are represented with 'X's, empty slots are represented with '.'s. You may assume the following rules: You receive a valid board, made of only battleships or empty slots. B…

一个新的方案,用的UBOOT 可能和上一个方案是同一个,但是配置有可能不一样,今天记录一下通过修改配置文件使用新的 board 文件以及 config.h 进入 u-boot 2011.09 // 打开 boards.cfg // vim boards.cfg // 这一行是旧的 171 ok335x arm armv7 ok335x forlinx ti81xx // 新增 ECM_5412 规则 172 ECM_5412 arm armv7 ECM_5412 aplex ti81xx //…

拿到一个uboot 后,我都想添加一个属于自己的board文件以及include/configs/*.h 文件. 如何添加这个些文件,今天来记录一下. 复制一份你所参考的板级文件,比如说board/vscom/baltos/ 复制为board/sbc7109 文件夹 修改board/sbc7109/Kconfig 里面的内容 if TARGET_AM335X_SBC7109 // 这个是等下make menuconfig 指定的一个宏 config SYS_BOARD //指定你的board 文…

Zybo Board开发记录: 让 Vivado有Zybo Board的配置文件 由 judyzhong 于 星期三, 07/13/2016 - 10:41 发表 本文转载自:coldnew's blog 在透过 Vivado 去建立新的项目的时候,开发板没有 Zybo Board 的选项可以选,我们就必须自己去设定关于 Zybo Board 的信息.幸好, Digilentinc 针对这个问题有提供关于 Zybo Board 的配置文件,就让我们来搞定他吧. 关于 Zybo Board 的配置…

1.项目结构 2.页面 (1)数据(逻辑) board.js // pages/board/board.js Page({ /** * 页面的初始数据 */ data: { imgWrap: [] }, /** * 生命周期函数--监听页面加载 */ onLoad: function (options) { const _this = this; // 请求数据 wx.request({ url: 'https://api.douban.com/v2/movie/coming_soon', da…

前言 原创文章,转载引用务必注明链接,水平有限,如有疏漏,欢迎指正. 本文使用Markdown写成,为获得更好的阅读体验和正常的链接.图片显示,请访问我的博客原文: http://www.cnblogs.com/sjqlwy/p/lattepanda_firmata.html 1.LattePanda与UP Board 由于Atom引出的GPIO接口较少,而且是1.8v电平,所以相关开发板都要使用额外的芯片进行拓展.胖哒板载一个Intel Atom z8350 处理器和实现Arduino 功能的…

Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be us…

Given a 2D board and a list of words from the dictionary, find all words in the board. Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same le…

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'. A region is captured by flipping all 'O's into 'X's in that surrounded region. For example, X X X X X O O X X X O X X O X X After running your function, the board should…

Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be us…

Write a program to solve a Sudoku puzzle by filling the empty cells. Empty cells are indicated by the character '.'. You may assume that there will be only one unique solution. A sudoku puzzle... ...and its solution numbers marked in red. 这道求解数独的题是在之…

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board could be partially filled, where empty cells are filled with the character '.'. A partially filled sudoku which is valid. Note:A valid Sudoku board (partially…

一直直到bug-free.不能错任何一点. 思路不清晰:刷两天. 做错了,刷一天. 直到bug-free.高亮,标红. 185,OA(YAMAXUN)--- (1) findFirstDuplicate string in a list of string. import java.util.HashSet; import java.util.Set; public class Solution { public static void main(String[] args) { String[…

------------------------------------------------------------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…

第18章---高度难题 1,-------另类加法.实现加法. 另类加法 参与人数:327时间限制:3秒空间限制:32768K 算法知识视频讲解 题目描述 请编写一个函数,将两个数字相加.不得使用+或其他算数运算符. 给定两个int A和B.请返回A+B的值 测试样例: 1,2 返回:3 答案和思路:xor是相加不进位.and得到每一个地方的进位.所以,用and<<1之后去与xor异或.不断递归. import java.util.*; public class UnusualAdd { pu…

#ifndef GAMEWIDGET_H #define GAMEWIDGET_H #include <QWidget> #include <QMouseEvent> #include <QEventLoop> #include <QTimer> #include <QPainter> #include <QList> // 手势的方向 enum GestureDirect { LEFT = , // 向左 RIGHT = , //…

题目简述: Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board could be partially filled, where empty cells are filled with the character '.'. A partially filled sudoku which is valid. Note: A valid Sudoku board (par…

Redis不是比较成熟的memcache或者Mysql的替代品,是对于大型互联网类应用在架构上很好的补充.现在有越来越多的应用也在纷纷基于Redis做架构的改造.首先简单公布一下Redis平台实际情况: 2200+亿 commands/day 5000亿Read/day 500亿Write/day 18TB+ Memory 500+ Servers in 6 IDC 2000+instances 应该是国内外比较大的Redis使用平台,今天主要从应用角度谈谈Redis服务平台. Redis使用场…

17.2 Design an algorithm to figure out if someone has won a game oftic-tac-toe. 这道题让我们判断玩家是否能赢井字棋游戏,有下面几点需要考虑: 1. 判断是否能赢hasWon函数是调用一次还是多次,如果是多次,我们可能为了优化而需要加入一些预处理. 2. 井字棋游戏通常是3x3的大小,我们是否想要实现NxN的大小? 3. 我们需要在代码紧凑,执行速度和代码清晰之间做出选择. #include <iostream> #…

在LeetCode中的Surrounded Regions 包围区域这道题中,我们发现用DFS方法中的最后一个条件必须是j > 1,如下面的红色字体所示,如果写成j > 0的话无法通过OJ,一直百思不得其解其中的原因,直到有网友告诉我说他验证了最后一个大集合在本地机子上可以通过,那么我也来验证看看吧. class Solution { public: void solve(vector<vector<char> >& board) { ; i < boar…

面试题17.1:编写一个函数,不用临时变量,直接交换两个数. 思路:使用差值或者异或 package cc150.middle; public class Exchange { public static void main(String[] args) { // TODO 自动生成的方法存根 int[] a= {1,2}; Exchange ec = new Exchange(); int[] b= ec.exchangeAB(a); System.out.println(b[0]); Sys…

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'. A region is captured by flipping all 'O's into 'X's in that surrounded region . For example, X X X X X O O X X X O X X O X X After running your function, the board should…