上一集我们说到如何从零开始搭建一个Vue-cli 3.0的项目,而这一集我们将说到如何手写一份阉割版的CI脚本. 首先说一下GitLab部署到服务器的操作,一般有两种,一种是规范化分离的,包含runner+ci+gitlab pages.另外一种是高自由度的,只包含runner+ci. 这两种部署方式的区别在于,前者的ci只包含关于打包和文件移动的操作(阉割版),至于部署到哪个服务器,具体怎么部署,一般来说是由项目组的管理员去配置的,开发者只需要把打包后的文件丢到一个名为public的目录,gi

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 这道题给我们一个二叉树,让我们返回所有根到叶节点的路径,跟之前那道Path Sum II 二叉树路径之和之二很类似,比那道稍微简单一

Given a binary tree, return all root-to-leaf paths. Note: A leaf is a node with no children. Example: Input: 1 / \ 2 3 \ 5 Output: ["1->2->5", "1->3"] Explanation: All root-to-leaf paths are: 1->2->5, 1->3 题目 给定一棵二叉树,

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t

题目: A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish'

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 给一个二叉树,返回所有根到叶节点的路径. Java: /** * Definition for a binary tree node

Given a binary tree, return all root-to-leaf paths.Example Given the following binary tree: 1 /   \2     3 \  5 All root-to-leaf paths are: [  "1->2->5",  "1->3"] LeetCode上的原题,请参见我之前的博客Binary Tree Paths. 解法一: class Solution {

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t

机器人位于一个 m x n 网格的左上角, 在下图中标记为“Start” (开始).机器人每次只能向下或者向右移动一步.机器人试图达到网格的右下角,在下图中标记为“Finish”(结束).问有多少条不同的路径? 详见:https://leetcode.com/problems/unique-paths/description/ Java实现: 参考:https://www.cnblogs.com/springfor/p/3886603.html class Solution { public i

一个机器人位于一个 m x n 网格的左上角 (起始点在下图中标记为"Start" ). 机器人每次只能向下或者向右移动一步.机器人试图达到网格的右下角(在下图中标记为"Finish"). 问总共有多少条不同的路径? 例如,上图是一个7 x 3 的网格.有多少可能的路径? 说明:m 和 n 的值均不超过 100. 示例 1: 输入: m = 3, n = 2 输出: 3 解释: 从左上角开始,总共有 3 条路径可以到达右下角. 1. 向右 -> 向右 ->

https://reacttraining.com/react-router/web/example/recursive-paths import React from 'react' import { BrowserRouter as Router, Route, Link } from 'react-router-dom' const PEEPS = [ { id: 0, name: 'Michelle', friends: [ 1, 2, 3 ] }, { id: 1, name: 'Se

网址:https://leetcode.com/problems/unique-paths/ 第一思路是动态规划 通过观察,每一个格子的路线数等于相邻的左方格子的路线数加上上方格子的路线数 于是我们就得到 dp[i][j] = dp[i-1][j] + dp[i][j-1] 这个状态转移方程 一开始通过调用函数的方式对dp数组赋值,并且每次都判断index是否为特殊值,感觉耗时会很多,果不其然! int get_sum(vector<vector<int>> dp, int i,

题目: A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish'

输入一颗二叉树和一个整数,打印出二叉树中结点值的和为输入整数的所有路径.路径定义为从树的根结点开始往下一直到叶结点所经过的结点形成一条路径. 本身题目不是很难,但是因为刚接触pyhon,对一些对象的传递不太了解,所以跳了很久也没有通过,后来看到这篇文章  后才明白,犯了一样的错误 # -*- coding:utf-8 -*- # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.ri

#!/usr/bin/env python # coding=utf-8 __author__ = 'Luzhuo' __date__ = '2017/5/7' import os def os_demo(): # 执行命令 dirs = os.popen("dir").read() print(dirs) # 打印目录树 dirs_info = os.scandir() for info in dirs_info: print("文件名: {}, 路径: {}, inode

1.requirejs遵循AMD规范,将需要的都加载好(前置加载).注:cmd是就近加载. define(['jQuery','dialog'],function($,d){ // 业务逻辑 }) (1)AMD采用异步加载,模块的加载不影响它后面语句的执行. (2)所有依赖这个模块的语句,都定义在一个回调函数中,等前面需要加载的模块加载完成后这个回调函数才运行,解决了依赖性的问题. 2.requirejs路径解析 代码结构为: 示例代码: index.html <!doctype html>

兼容windows和linux. 分析: 在windows下路径有以下表示方式: (标准)D:\test\1.txt (不标准,参考linux)D:/test/1.txt 然后在java中,尤其使用File对象在操作windows的路径时,非常的不好操作,因为不知道要用多少个斜杠“\”来表示. 斜杠“\”为转义字符(参考:http://www.cnblogs.com/EasonJim/p/6561576.html). 而java的File对象上无论是windows还是linux,输入的路径不管标

一.path核心模块 ①path.basename(path[,ext])获取一个路径中的文件名 var path=require('path'); console.log(path.basename('C:/Files/nodejs/a.js'));//a.js console.log(path.basename('C:/Files/nodejs/a.js','.js'));//a ②path.dirname(path)  获取一个路径中的目录 var path=require('path')

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1 return true

Hdfs的访问方式有两种,第一:类似linux命令,hadoop shell.第二:java API方式. 先看第一种. FS Shell cat chgrp chmod chown copyFromLocal copyToLocal cp du dus expunge get getmerge ls lsr mkdir movefromLocal mv put rm rmr setrep stat tail test text touchz FS Shell 调用文件系统(FS)Shell命令