首先,先通过以下函数向BookList集合中插入10000条数据 function insertN(obj,n){var i=0;while(i<n){obj.insert({id:i,name:"bookNumber"+i,publishTime:i+2000})i++;}}var BookList=db.getCollection("BookList")调用函数,这样,BookList
Description Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this
1.场景 假设有一个user用户表,表字段分别为:id(主键),name(姓名),age(年龄),hobby(爱好)。而一个人可能有好几个爱好,游泳啊篮球啊乒乓球啊等等。数据库里hobby字段存的是:游泳,篮球,乒乓球 而要想查所有喜欢游泳的人,就可以用find_in_set函数了 todo:贴图 2.使用 select *from user where find_in_set("游泳"
使用 ViewPagerIndicator 时 , 布局文件报的错 Missing styles. Is the correct theme chosen for this layout? Use the Theme combo box above the layout to choose a different layout, or fix the theme style refe
题:https://leetcode.com/problems/find-all-anagrams-in-a-string/description/ 题目 Given a string s and a non-empty string p, find all the start indices of p’s anagrams in s. Strings consists of lowerca
传送门:【HDU】1595 find the longest of the shortest 题目分析:首先求出一条最短路,记录下最短路上用到的边,枚举删除每一条边,求一次最短路,求完后恢复删除的边。重复这一过程直到枚举完所有的边为止。所有删除边后求得的最短路里最长的那条就是答案。 代码如下: #include <cstdio>#include <cstring>
Union Find模板要会背诵; private class UnionFind {private int[] father;private int count;public UnionFind(int n) {this.father = new int[n + 1];for(int i = 0; i <= n; i++) {father[i] = i;}this.count = n;}pub
首先quick select算法的模板要倒背如流,这个是通过quick sort里面提炼得到的算法;两个while一个if,condition相同;后面再递归 Kth Largest Element in an Array (是找从大到小的第k大;注意左边是大的,右边是小的,quick select的模板要熟记) class Solution {public int findKthLarges
比如:查找当前路径下,所有以path结尾的路径,然后查看这些路径下是否含有以transcript.fa.fa结尾的文件,如果有,删除 #!/bin/bash# 获取当前路径下以rG4结尾的所有目录dirs=$(find . -type d -name "*path")# 遍历每个以rG4结尾的目录for dir in $dirs; do# 查找匹配的文件files=$(find "$dir"