ICode9

NumPy 教程目录 Lesson11——NumPy 位运算 　　NumPy "bitwise_" 开头的函数是位运算函数。 　　NumPy 位运算包括以下几个函数： 函数描述 bitwise_and 对数组元素执行位与操作 bitwise_or 对数组元素执行位或操作 bitwise_xor 对数组元素执行位异或操作 invert 按位

The problem of this solution can be BFS and DFS. BFS is easy to understand. The following is DFS solution: private List<Integer> res = new ArrayList<>(); public List<Integer> rightSideView(TreeNode root) { helper(root, 0

马上就要考 SCP 了，赶紧写一些题解积累人品 P3855 [TJOI2008]Binary Land 题解 很明显 可以看出这道题就是一个 BFS 的题目，利用 BFS 暴力搜到答案。 题目在讲什么 先输入行数 \(N\) 列数 \(M\)，再输入一个 \(N * M\) 的字符串矩阵 有 1 对情侣，他们的位置在矩阵中分别用 \(G\), \(M\)

在使用静态变量Map作为返回时，抛出了异常 com.alibaba.fastjson.JSONException: write javaBean error, fastjson version 1.2.76, class org.apache.flink.table.data.binary.BinaryStringData, fieldName : id, Memory segment does not represent off heap memory 其余代

python解析pb二进制文件，写入明文文本 背景: 在项目中需要解析pb二进制文件，转为明文，写入txt文本中保存，同时转为由分隔符进行分隔的行列式结构，在这记录一下这个过程，以下列出了两种方法：方法二存在bytes解析失败的问题，采取方法一： 主要由以下四步组成： 1、二进制数据反序列化 2、反序列

一、获取File对象 // 获取File对象 let firstImg = data.add_design.first_pic_whether.raw; 1、 用户上传的图片数据 .1.1、 图片数据详情 二、使用new formData() let binary = new FormData() binary.append('file',firstImg) binary.append('name','2') 三

The first solution of this problem can be based on the 236. Lowest Common Ancestor of a Binary Tree too: public Node lowestCommonAncestor(Node p, Node q) { Node root = p; while(root.parent!=null) root = root.parent;

This problem is just as same as "235. Lowest Common Ancestor of a Binary Search Tree", the only difference is, 235 is two points, 1676 is 1~n points. The solution is almost same, the only difference is in this solution, we use a set to check whe

When we get this problem, we need to confirm the following 2 questions: 1. Can root, p or q be null? (No) 2. Are both p and q in the tree (No, either p or q mignt not in the tree), this is the only difference with 236. Lowest Common Ancestor of a Binary T

When we get this problem, we need to confirm the following 2 questions: 1. Can root, p or q be null? (No) 2. Can p be equal to q? (No) We look "root" as a pointer, the point will check recursively of it's left and right sub-tree. If the lef

首先遇到这个是因为binlog位置索引处的问题，不要reset slave； reset slave会将主从同步的文件以及位置恢复到初始状态，一开始没有数据还好，有数据的话，相当于重新开始同步，可能会出现一些问题； 一般做主从同步，都是要求以后的数据实现主从同步，而对于旧的数据完全可以使用数据库同步工具先

 题目大意:已知一个树的前序和中序序列，求两个结点的最近公共祖先 本着试一试的做法，直接建树，然后套倍增求LCA的算法，还好数据不是很大，没卡爆数组。 #include <iostream> #include <unordered_set> #include <cstring> #include <cmath> using namespace std; const int N = 100

For this problem, if don't consider the follow up limitation, the solution is very easy: The time complexity is O(1), the space complexity is O(n), n is the total node number of the tree. class BSTIterator { List<Integer> list = new Array

安装 sudo apt-get install python-capstone git clone https://github.com/JonathanSalwan/ROPgadget.git cd ROPgadget sudo python setup.py install 运行 m1sceden4@DESKTOP-H37I3CV:~/ROPgadget$ ROPgadget 报错 运行之后可能会发现报错 Traceback (most recent cal

This problem should not use DFS to solve it, I have tried, but failed with this test case: [3,9,8,4,0,1,7,null,null,null,2,5] My output is: [[4],[9,5],[3,0,1],[2,8],[7]] But it was expected: [[4],[9,5],[3,0,1],[8,2],[7]] Because the problem's request

This is a Post Order binary tree problem. For every node, we need to know its left tree's maximal diameter and its right tree's maximal diameter, and add left and right, we can get a sub-tree's maximal diameter, we just find the maximum of

This is a "In Order" traversal problem, because the order of the result is a in order or the tree. 1. The problem need to return the pointer to the smallest element of the linked list, so we first need to find the smallest element. 2. The proble

I have originally write a BFS solution for this problem, but it seems too difficlut. And then I read other's code, found that solution is much easier. The solution: 1. find all left boundary nodes 2. find all left leaves 3. find all right leaves 4. f

First I wrote a bruteforce solution, but it didn't pass because of too many calls, the time complexity is O(m*n): public int leftMostColumnWithOne(BinaryMatrix binaryMatrix) { List<Integer> dimensions = binaryMatrix.dimensions();

文章目录 前言1110 Complete Binary Tree (25 分)分析AC代码简化 总结 前言 今日完成任务： 1、单词150旧+50新 2、英语阅读手译一篇 3、英语练字 4、Java核心技术（回顾基础篇）（1.30h） 5、springBoot框架回顾（1.30h） 6、阅读毕设参考材料 7、Pat一题（1110） 提示：以下是本篇文章

题目链接 2种方法 结论版 一个很显然的结论就是\(n\)个数里两两之间最小异或和一定是相邻的\(2\)个数 于是就有了以下\(DP\)， 先将原数列排序 令\(f[i]\)表示最大的数为\(a_i\)的情况下最多能选几个数 显然有以下转移 \[f[i]=\max_{a_j\xor a_i \neq k} {f[j]} + 1 \]

正向思维硬解得分28，最初不知道中序遍历如何解，测试点3不思考为啥错误了 #include <cstdio> #include <algorithm> #include <cmath> #include <queue> #include <vector> using namespace std; const int maxn = 1010; int orinOrder[maxn],N; vector<int> orderCBT,le

题目描述 Given an integer n, return the number of structurally unique BST’s (binary search trees) which has exactly n nodes of unique values from 1 to n. 解题思路 动态规划。找动态规划方程，我们知道当加入第 i 个节点时，他的位置只能为：(左上) -> i -> （左下）。那么

1 . 二叉树的遍历 1020 Tree Traversals 1086 Tree Traversals Again 1102 Invert a Binary Tree 2 . 树的遍历 1090 Highest Price in Supply Chain 1106 Lowest Price in Supply Chain 1079 Total Sales of Supply Chain 1094 The Largest Generation 1004 Counting Leaves