cracking专题
Cracking the Safe
原题链接：https://leetcode.cn/problems/cracking-the-safe/description/ 题目要求的是，某个时刻能够打开保险箱的任一最短密码序列，需要包含所有密码子串。 答案应当是一个字符串，任意长度为n的子串的都是一种密码方案。 对于有n位，每位k种方案的密码串，共有k^n个。 题目要求最短，那么任意位置选出的子串应当是不重复的。 也就是说，一个
{ Cracking The Coding Interview: 150 programming QA } --- Arrays and Strings
Hashtable, ArrayList (Dynamically resizing array, providing O(1) access), StringBuffer (do string concatenation) 1. 判断string是否有重复字符，不允许使用其他数据结构。 Step 1: 问清楚限制，string的编码是ASCII还是Unicode a. 如果可以用其他数
Cracking Public keys
Public keys I have generated public keys for each of you. You can download the public keys here: public-keys.tgz (hps:/umd.instructure.com/courses/1358965/files/78500682?wrap=1) (hps:/umd.instructure
【网络攻防实验】【北京航空航天大学】【实验三、口令破解（Password Cracking）实验】
实验三、口令破解（Password Cracking）实验 一、 L0phtCrack破解实验 1、 注册L0phtCrack： 2、 设置口令： （1） 创建3个新账户： 帐户创建过程（以test-1为例）： 帐户创建结果： （2） 使用L0phtCrack破解口令：（使用管理员账号运行程序） 口令破解结果： 正确破解口令。
Cracking The Coding Interview 3.2
//How would you design a stack which, in addition to push and pop, also has a function min which returns the minimum element? Push, pop and min should all operate in O(1) time.//使用一个链表来记录最小值的index
Cracking The Coding Interview 3.2
//How would you design a stack which, in addition to push and pop, also has a function min which returns the minimum element? Push, pop and min should all operate in O(1) time.//使用一个链表来记录最小值的index
Cracking The Coding Interview 2.5
这题的思想来自于http://hawstein.com/posts/2.5.html，重新实现了一下 用hash来记录循环的起点 //Given a circular linked list, implement an algorithm which returns node at the beginning of the loop.//DEFINITION//Circula
Cracking The Coding Interview2.4
删除前面的linklist，使用node来表示链表 // You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1’s digit is at the head
Cracking The Coding Interview2.3
#include <iostream>#include <string>using namespace std;class linklist{private:class node{public:node(){}string data;node * next;};int size;public:node *head;linklist(){head = new node;size = 0;
Cracking The Coding Interview2.3
#include <iostream>#include <string>using namespace std;class linklist{private:class node{public:node(){}string data;node * next;};int size;public:node *head;linklist(){head = new node;size = 0;
Cracking The Coding Interview 2.2
#include <iostream>#include <string>using namespace std;class linklist{private:class node{public:node(){}string data;node * next;};node *head;int size;public:linklist(){head = new node;size = 0;
Cracking The Coding Interview 2.0 单链表
#include <iostream>#include <string>using namespace std;class linklist{private:class node{public:node(){}string data;node * next;};node *first;int size;public:linklist(){first = new node;size =
Cracking The Coding Interview 2.0 单链表
#include <iostream>#include <string>using namespace std;class linklist{private:class node{public:node(){}string data;node * next;};node *first;int size;public:linklist(){first = new node;size =
Cracking The Coding Interview 1.8
//Assume you have a method isSubstring which checks if one word is a substring of another. //Given two strings, s1 and s2, write code to check if s2 is a rotation of s1 using only one call to isSubst
Cracking The Coding Interview 1.8
//Assume you have a method isSubstring which checks if one word is a substring of another. //Given two strings, s1 and s2, write code to check if s2 is a rotation of s1 using only one call to isSubst
Cracking The Coding Interview 1.7
//Write an algorithm such that if an element in an MxN matrix is 0, its entire row and column is set to 0.////这题原答案就是要两个buffer来记录出现0的位置。不知道有没可以不用buffer的方法。#include <iostream>using namespace std;vo
Cracking The Coding Interview 1.7
//Write an algorithm such that if an element in an MxN matrix is 0, its entire row and column is set to 0.////这题原答案就是要两个buffer来记录出现0的位置。不知道有没可以不用buffer的方法。#include <iostream>using namespace std;vo
Cracking The Coding Interview 9.7
//原文：//// A circus is designing a tower routine consisting of people standing atop one another’s shoulders. For practical and aesthetic reasons, each person must be both shorter and lighter than the
Cracking The Coding Interview 9.6
//原文：//// Given a matrix in which each row and each column is sorted, write a method to find an element in it.// 从最右一排，从上到下，小于右上角的元素则下降，大于右上角的元素则左移找；第一个相等的元素。#include <iostream>using namespace s
Pycharm 2021.2.1 Cracking
Pycharm Cracking 写在文前：该Cracking方法，最高只适用Pycharm 2021.2.1 ， 2021.2.1以下版本笔者没试过，有兴趣可尝试。 2021.2.1以上版本皆不适用（笔者皆亲试），请另寻它法。 先下载Cracking工具： 百度网盘链接：https://pan.baidu.com/s/1xROaIC5NDKbL3NVhu6Nofg?pwd=bxii 提取
Cracking C++(6): 准确打印浮点数
文章目录 1. 目的2. 准确打印浮点数： 使用 fmt 库3. 准确算出被表示的值3.1 直观感受IEEE-754: float-toy3.2 获取浮点数二进制表示3.3 float 类型3.4 double 类型3.5 fp16 类型3.6 验证 4. 结论和讨论5. References 1. 目的 给 float 或 double 类型的变量赋值后， 打印出来的值和赋值