计算机科学(哈佛cs50)笔记汇总

Posted 2022-05-02 mask哥

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1 c语言

程序编译系统构成：预处理器、编译器、汇编器、连接器
代码执行的过程：源代码->编译器->汇编代码->汇编器->目标代码->连接器->可执行程序
程序的构成：输入+算法+程序设计+输出
ascill （https://www.asciichart.com/）
Unicode 指定标准语言
Cs50 编程手册：https://manual.cs50.io/
Pseudo code 伪代码
程序调试的方法：
1.控制台打日志printf
2.debugger
3.rubber duck-头脑风暴与同伴讨论代码
动态内存分配技术，自动扩大或缩小阵列
提倡使用快捷键、shell命令编程，提高开发效率
c语言非零即真

2 算法：

running time: 大O
divide and conquer strategy:分而治之算法
Linear search :线性搜索
Binary search 二分查找，数组是有序的，从中间查找目标数字，左边数字或右边数字与目标数字比对。直到找打目标数据。
Selectiton sort:选择排序
merge sort :合并排序
bubble sort: 冒泡排序
算法涉及对数

3 内存

指针
结构体:struct
strcmp:字符串比较
Dynamic Memory：动态内存分配
malloc :内存分配
Free :释放内存
值类型交换，变量值不变；引用类型交换，变量值改变。
Scanf(“%i”,&x) ：从键盘输入
c语言编译：make xx.c 执行 ./xx

4 数据结构

struct
Linked list:链表

#include <stdio.h>

#include <stdlib.h>

/**

*link list实现

typedef struct node

int number;

struct node *next;

node;

int main(void)

// list of size

node *list = NULL;

// add a nuber to list
node *n = malloc(sizeof(node));

if (n == NULL)

return 1;

n->number = 1;

n->next = NULL;

// update list to point to new node
list = n;

// add a number to list

n = malloc(sizeof(node));

if (n == NULL)

free(list);

return 1;

n->number = 2;

n->next = NULL;

list->next = n;

// add a nuber to list

n = malloc(sizeof(node));

if (n == NULL)

free(list->next);

free(list);

return 1;

n->number = 3;

n->next = NULL;

list->next->next = n;

// print numbers

// for(node *tmp =list;tmp!=NULL;tmp=tmp->next)

// printf(“%i\\n”,tmp->next);

// free list

while (list != NULL)

node *tmp = list->next;

free(list);

list = tmp;

return 0;

Tree
Binary search tree 二叉搜索树；根节点大于左子节点，小于右子节点

typedef struct node

int number;

struct node *left;

struct node *right;

node;

void free_tree(node *root);

void print_tree(node *root);

int main(void)

// tree of the size 0

node *tree = NULL;

// add number to list

node *n = malloc(sizeof(node));

if (n == NULL)

// free the memory

return 1;

n->number = 1;

n->left = NULL;

tree = n;

// add number to list

n = malloc(sizeof(node));

if (n == NULL)

return 1;

n->number = 1;

n->left = NULL;

n->right = NULL;

tree->left = n;

// print tree

print_tree(tree);

// free tree

free_tree(tree);

void print_tree(node *root)

if (root == NULL)

return;

print_tree(root->left);

printf(“%i\\n”, root->number);

print_tree(root->right);

free_tree(root);

/**

*** @brief

** 二分查找树*

bool search(node *tree, int number)

if (tree == NULL)

return 1;

else if (number < tree->number)

return search(tree->left, number);

else if (number > tree->number)

return search(tree->right, number);

else if (number == tree->number)

return 0;

AVL或红黑树
Hash table:哈希表=数组+链表

hash冲撞->散列函数
Tries :恒定的时间查找，对于海量数据集也是；就是多层哈希表；由树和数组组成。
Graph
Queues 队列先进先出(FIFO)
Enqueue/dequeue
Stack 后进先出 LIFO
push/pop
Abstract data structures

5 python

python中定义的main函数，必须在程序最后一行调用。
docs.python.org
google & stackoverflow 检索问题
Pip python包安装管理，类似java中maven/gradle

pip install xx
命令行携带参数

from sys import argv

#命令行携带参数

if len(argv) == 2:

print(f"hello,argv[1]")

else:

print(“hello,world”)

执行运行结果：
with关键字使用
python 数据分析
python语音处理
python 人脸识别
face_recoginition模块
微服务springcloud/springcloudalibaba sidecar 调用python api
python生成二维码（QR）
python 代码范例：

agree.py
s = input(" Do you agree? ")
print("s:"+s)
s.lower()
if s == "Y" or s == "y":
    print("Agree")
elif s == "N" or s == "n":
    print("No agreement")

argv.py
from sys import argv
# 命令行携带参数
# if len(argv) == 2:
#     print(f"hello,argv[1]")
# else:
#     print("hello,world")

for arg in argv[2:]:
    print(arg)

blur.py
from PIL import Image, ImageFilter

# 图片滤镜
before = Image.open("vscode.jpg")
# after = before.filter(ImageFilter.CONTOUR)
after = before.filter(ImageFilter.BoxBlur(10))
after.save("cp2vscode.jpg")

caculator.py

try:
    x = int(input("x: "))
    y = int(input("y: "))
    print(x/y)
except ValueError:
    print("that is not an int")
    exit()

crackPwd.py
from string import ascii_letters, digits, punctuation
from itertools import product
# 暴力破解
for passcode in product(ascii_letters + digits + punctuation, repeat=8):
    print(passcode)

dic.py
# 字典 hashtable set 去重
words = set()
def check(word):
    if word.lower() in words:
        return True
    else:
        return False
def load(dictionary):
    file = open(dictionary, "r")
    for line in file:
        word = line.rstrip()  # /n分割单词
        words.add(word)
    file.close()
    return True
def size():
    return len(words)
def unload():  # 释放内存
    return True

edges.py
from PIL import Image, ImageFilter
# 滤镜处理
before = Image.open("vscode.jpg")
after = before.filter(ImageFilter.FIND_EDGES)
after.save("cp3vscode.jpg")

mario.py
print("?"*4)

for i in range(4):
    print("?"+str(i), end="$")  # 使用$进行分割字符串

meow.py
def main():
    for i in range(3):
        meow(8)
def meow(n):
    print("meow"+str(n))
main()  # python中必须调用定义的入口函数main

phonebook.py
# 数据结构hashtable&线性查找
people = 
    "mark": "110",
    "steve": "112"


name = input("name: ")
if name in people:
    num = people[name]
    print(f"Number:num")

qrcode.py
import os
import qrcode
#二维码生成
img = qrcode.make(
    "https://www.baidu.com/img/PCtm_d9c8750bed0b3c7d089fa7d55720d6cf.png")
img.save("qr.png", "PNG")
os.system("open qr.png")  # 调用系统命令open 打开二维码

scores.py
scores = [33, 55, 75]
# 求和
avg = sum(scores) / len(scores)
# print("avg"+str(avg))
print(f"avg:avg")

scores2.py
scores = []
for i in range(3):
    score = int(input("score: "))
    # scores.append(score)
    scores += [score]

avg = sum(scores) / len(scores)
print(f"avg:avg")