Year: 2022

之前看人推荐logseq就尝试了一下，吐槽跨设备同步难搞，然后被人种草了Obsidian，这几天使用下来感觉确实比logseq好用多了，第三方插件同步虽然有点冲突但大体上还好。现在考虑怎么去合理规划tag以及wiznote笔记的迁移，虽说感觉也不能完全替代wiznote。这个还是比较适合纯文本记录，对文件等等同步不太友好。

最近想给一个APP添加分享屏幕截图的功能，就尝试找了一圈相关的Package，最后找倒是没找到，倒是在StackOverflow找到了一个解决方法。

最近尝试学了一下K8S，感觉还行吧！只是对于我自己的项目来说还没必要，弄个docker就不错了，集群对于小项目来说部署有点夸张。个人看了这几天发现官方的交互式教程还是挺爽的，简单了解还是可行的。我这就弄个从头到尾的单节点部署让大家了解一下，可能稍微有点乱。

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string. Note: You must not use any built-in BigInteger library or convert the inputs to integer directly.【The sadest things which I can found the fastest soultion in leetcode break this rule.】

最近学习这个k8s在PVC存储这儿卡了两天，今天终于算是大概搞明白是个怎么回事了。 以下是该过程的总结： 创建由物理存储支持的 PersistentVolume。你不会将卷与任何 Pod 关联。 创建一个 PersistentVolumeClaim， 它将自动绑定到合适的 PersistentVolume。 创建一个使用 PersistentVolumeClaim 作为存储的 deployment

You are given two lists of closed intervals, firstList and secondList, where firstList[i] = [starti, endi] and secondList[j] = [startj, endj]. Each list of intervals is pairwise disjoint and in sorted order. Return the intersection of these two interval lists. A closed interval [a, b] (with a

Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Given an array intervals where intervals[i] = [li, ri] represent the interval [li, ri), remove all intervals that are covered by another interval in the list. The interval [a, b) is covered by the interval [c, d) if and only if c

Alice and Bob play a game with piles of stones. There are an even number of piles arranged in a row, and each pile has a positive integer number of stones piles

You are playing the following Nim Game with your friend: Initially, there is a heap of stones on the table. You and your friend will alternate taking turns, and you go first. On each turn, the person whose turn it is will remove 1 to 3 stones from the heap. The one who removes the…