songbo


项目简介- MWC 红板飞控 主控芯片MEGA2560

自己的毕业设计是四轴飞行器的手机控制,需要学习Arduino来编写飞控软件。因此写一系列Arduino开发笔记。一方面帮助自己理解技术知识,一方面帮助可能与我遇到相同问题的人。 我所用的硬件:MWC红板飞控,搭载MEGA2560芯片,板载三轴角速度传感器ITG3205(ITG3200),三轴加速度传感器BMA180,三轴磁罗盘HMC5883,气压计BMP085(9D0F)。


[leetcode] Multiply Strings

Multiply Strings Given two numbers represented as strings, return multiplication of the numbers as a string. Note: The numbers can be arbitrarily large and are non-negative. Tags: math, string 字符串相乘,关键是模拟我们平时手算乘法时的过程。取出num2的每一位,乘以num1,再把之前得到的re向左移一位,再加上这个循环中的结果。所以我们需要实现两个函数multiplyNum(字符串乘以数字)和sum(两个字符串相加)。 num1或num2为零需要单独考虑,否则结果会出现类似0000的情况。 class Solution { public: string multiply(string num1, string num2) { int i = 0; string re; if(num1 == “0” […]


[leetcode] Combinations

Combinations Given two integers n and k, return all possible combinations of k numbers out of 1 … n. For example, If n = 4 and k = 2, a solution is: [ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4], ] tag: backtracking 回溯问题,用递归求解就好。 class Solution { public: vector<vector<int> > combine(int […]


[leetcode] Longest Valid Parentheses

Longest Valid Parentheses Given a string containing just the characters ‘(‘ and ‘)’, find the length of the longest valid (well-formed) parentheses substring. For “(()”, the longest valid parentheses substring is “()”, which has length = 2. Another example is “)()())”, where the longest valid parentheses substring is “()()”, which […]


[leetcode] Next Permutation

Next Permutation Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). The replacement must be in-place, do not allocate extra memory. Here are some examples. […]


[leetcode] N-Queens II

N-Queens II Follow up for N-Queens problem. Now, instead outputting board configurations, return the total number of distinct solutions. tag: backtracking 此题与上一题非常相似,只是把“输出所有可能情况”改成了“输出所有可能情况的个数”。详细解析请看上一题的结题报告: N-Queens class Solution { public: int totalNQueens(int n) { if(n == 0){ return 0; } int queen[n] = {0}; int sum = 0; for(int i = 0; i < […]