LouisLan的学习简记 👋🏼

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安卓接入高德地图API展示热力图 概述：因为要用到这个，所以记一下怎么弄的，以后可能会用到。废话不多说； 首先，去高德地图 API 官网搞一个 key。 KEY 名称自定义，中英文都可以，只是让你分辨你的key， 安全码 SHA1 的获取见： https://www.lonuslan.com/articles/2020/06/23/1592905144300.html 建立好后会在我的应用里生成一个 KEY ，记住这个，复制下这个后面会用到。 之后在 AndroidManifest.xml 中 <application 标签内添加： <!-- //配置应用的AppKey,唯一的，需申请 --> <meta-data android:name="com.amap.api.v2.apikey" android:value="您刚刚申请的key"/> <!-- 声明定位 service--> <service android:name="com.amap.api.location.APSService"/> 在 <manifes....

安卓接入企业微信 概述：博主是个三流代码搬运工，最近接手某个课题，需要做一个数据可视化的 app ，将一些信息直观的统计并且展示在地图上，所以也会用到高德地图 API 的接口，所以后续也会分享一下使用高德 API 遇到的一些问题。有技术的看个热闹，没技术的窥个门道，权当增加业余知识~~ 准备工作 既然要接入企业微信，必要的企业 微信管理员账号 还是要有的！ 首先建一个类保存一下登陆时可能需要用到的各种 key ，具体看图片中每个 key 所对应的值； QYWXConstants.class public class QYWXConstants { //企业id public final static String CORPID = "***"; //应用ID public final static String AGENT_ID = "***"; //应用密钥 public final static String SECRET = "***"; //应用schema public final static String SCHEMA = "***"; //签名的key pub....

题目地址 https://leetcode.com/problems/triangle/ 题目描述 Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. For example, given the following triangle [ [2], [3,4], [6,5,7], [4,1,8,3] ] The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11). Note: Bonus point if you are able to do this using only O ( n ) extra space, where n is the total number of rows in the triangle. 思路 这道题给了我们一个二维数组组成的三角形，让我们寻找一条自上而下的路径，使得路径和....

题目地址 https://leetcode.com/problems/pascals-triangle-ii/ 题目描述 Given a non-negative index k where k ≤ 33, return the k th index row of the Pascal's triangle. Note that the row index starts from 0. In Pascal's triangle, each number is the sum of the two numbers directly above it. Example: Input: 3 Output: [1,3,3,1] Follow up: Could you optimize your algorithm to use only O ( k ) extra space? 思路 杨辉三角想必大家并不陌生，应该最早出现在初高中的数学中，其实就是二项式系数的一种写法。 　　　　　　　　１ 　　　　　　　１　１ 　　　　　　１　２　１ 　　　　　１　３　３　１ 　　　　１　４　６　４　１....

题目地址 https://leetcode.com/problems/pascals-triangle/ 题目描述 Given a non-negative integer numRows , generate the first numRows of Pascal's triangle. In Pascal's triangle, each number is the sum of the two numbers directly above it. Example: Input: 5 Output: [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] 思路 杨辉三角是二项式系数的一种写法，如果熟悉杨辉三角的五个性质，那么很好生成，可参见我的上一篇博文Pascal's Triangle II。具体生成算是：每一行的首个和结尾一个数字都是 1，从第三行开始，中间的每个数字都是上一行的左右两个数字之和。 关键点解析 代码 class Solution { public: vector<vector<int>> gen....

题目地址 https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/ 题目描述 Given a binary tree struct Node { int val; Node *left; Node *right; Node *next; } Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL. Initially, all next pointers are set to NULL. Example: Input: root = [1,2,3,4,5,null,7] Output: [1,#,2,3,#,4,5,7,#] Explanation: Given the above binary tree (Figure A), your function should popula....

题目地址 https://leetcode.com/problems/populating-next-right-pointers-in-each-node/ 题目描述 You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition: struct Node { int val; Node *left; Node *right; Node *next; } Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL. Initially, all next pointers are set to NULL. Follow up: You m....

题目地址 https://leetcode.com/problems/distinct-subsequences/ 题目描述 Given a string S and a string T, count the number of distinct subsequences of S which equals T. A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not). It's guaranteed the answer fits on a 32-bit signed integer. example1: Input: S....

题目地址 https://leetcode.com/problems/flatten-binary-tree-to-linked-list/ 题目描述 Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 click to show hints. Hints: If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order trave 思路 这道题要求把二叉树展开成链表，根据展开后形成的链表的顺序分析出是使用先序遍历，那么只要是数的遍历就有递归和非递归的两种方法来求解，这里我们也用两种方法来求解。首先来看递归版本的，思路是先利用 DFS 的思路找到最左子节点，然后回....