https://github.com/xiantang/grokking_algorithms
grokking algorithms 《图解算法》代码
https://github.com/xiantang/grokking_algorithms
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grokking algorithms 《图解算法》代码
- Host: GitHub
- URL: https://github.com/xiantang/grokking_algorithms
- Owner: xiantang
- Created: 2018-10-13T13:59:10.000Z (over 6 years ago)
- Default Branch: master
- Last Pushed: 2022-01-25T04:25:12.000Z (over 3 years ago)
- Last Synced: 2025-04-06T08:11:17.808Z (3 months ago)
- Topics: algorithms, grokking-algorithms
- Language: Python
- Homepage:
- Size: 21.5 KB
- Stars: 17
- Watchers: 0
- Forks: 6
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# 图解算法代码
这是本人在18年学习《图解算法》时候学习留下的代码和相应的思考，这本书是一本很不错的算法书籍，如果认真阅读思考可以为你的算法打下不错的基础，在阅读了这本书之后可以去阅读《算法4th》《算法导论》等书，同时也可以辅以 leetcode 练习实践一下。
对于这个仓库的问题和疑问欢迎在 issue 区讨论或者提出 MR 修改代码👏。
### 二分查找
```python
import math
def bina_search(list,item):
low = 0
hight = len(list)-1
while low<=hight:
#每次猜测
mid = math.floor((low + hight)/2)
guess = list[mid]
if guess == item:
return mid
if guess > item:
hight = mid-1
else:
low = mid+1
return None
my_list = [1,3,5,7,8]
print(bina_search(my_list,7))
```
这个思路主要是不断中断数组
然后判断是否在中点上
* 疑问: 为什么判断错误之后 mid 要+1/-1呢?
* 回答: 画了张图理解了一下 主要是把猜错的中点给排除掉
### 选择排序
```python
def find_small(arry:list)->int:
"""
找到最小的元素
:param arry:
:return:
"""
smallest = arry[0]
smallest_index = 0
for i in range(1,len(arry)):
if smallest>arry[i]:
smallest = arry[i]
smallest_index = i
return smallest_index
def selectionSort(arry:list)->list:
"""
选择排序
:param arry:
:return:
"""
new_array = []
for i in range(len(arry)):
smallest = find_small(arry)
new_array.append(arry.pop(smallest))
return new_array
print(selectionSort([5,3,6,2,10]))
```
* 疑问:为什么循环`len(array)`次就可以完成了排序?
* 回答:因为数组长度是`len(array)`,每次都会弹出一个元素,
弹`len(array)`次就会让原数组为空.
### 数组链表区别
* 数组元素是在一起的/列表是分开的,每个元素到都存储了下一个元素的地址
* 数组读取,随机读取很快/链表插入删除很快.
* 数组中所有元素都是一个类型.
### 递归
```python
def countdown1(i):
print(i)
countdown1(i-1)
# countdown(1)
def countdown2(i):
print(i)
if i<=1:#基线条件
return
else: #递归条件
countdown2(i-1)
a=countdown2(3)
```
递归主要由两个部分组成:
1. 递归条件:函数调用自己
2. 基线条件:函数不再调用自己
### 调用栈
* 当调用另一个函数时候,当前函数处于未完成状态.
```python
def fact(x):
if x==1:
return 1
else:
return x*fact(x-1)#塞入栈中 #fact函数没有结束
```
### 递归总结
* 递归指的是调用自己的函数
* 1. 递归条件:函数调用自己
2. 基线条件:函数不再调用自己
* 两种操作1.压入栈中 2.弹出栈
* 所有函数都进入调用栈
* 调用栈很长就会很占用内存
### D&C算法
```python
def DC(x,y):
if x>y:
x=x%y
if x ==0:#基线条件
return y
return DC(x,y)
if x符合基线条件
#### 求和
```python
#我写的
a = [1,2,3,4,5]
def sum(array):
if len(array) ==0 :
return 0
if len(array) ==1 :
return array[0]
else:
first = array[0]
array.pop(0) #pop 指的是pop的索引
return first + sum(array)
print(sum(a))
```
```python
#标准答案
a = [1,2,3,4,5]
def sum(array):
if array == []:
return 0
else:
return array[0] + sum(array[1:])
print(sum(a))
```
数组求和问题
1. 将数组问题分割成数组第一个元素之后的元素和第一个元素相加
2. 涉及数组的递归函数时候,基线条件通常是数组为空/只包含一个元素
#### 求数目
```python
a = [1,2,3,4,5]
def count(array):
if array == []:
return 0
else:
return 1+count(array[1:])
print(count(a))
```
### 求最大值
```python
a = [1, 2, 8, 4, 5]
def biggest(array):
if array == []:
return 0
else:
return array[0] if\
array[0] > biggest(array[1:])\
else biggest(array[1:])
print(biggest(a))
```
* 或许我们接触的是for循环/如果我们用递归呢?
使用递归实现二分法
```python
import math
a = [1, 2, 3, 4, 5, 8, 10]
def cut(array, item):
len_array = len(array)
if len_array == 1:
return 0
else:
mid = math.floor(len_array / 2)
if item == array[mid]:
return mid
if item > array[mid]:
return len(array[:mid]) + cut(array[mid:], item)
else:
return cut(array[:mid], item)
print(cut(a, 8))
```
* 基线条件
1. 只剩下一个元素,就是那个需要的元素
2. 中点直接是猜到的元素
* 递归条件
1. 猜的元素比中点大
2. 猜的元素比中点小
### qsort
```python
a = [1,2,6,3,8,5]
def qsort(array):
if len(array) < 2:
return array
mid = array[0]
low = [array[i] for i in range(1,len(array)) if array[i] <= mid]
up = [array[i] for i in range(1,len(array)) if array[i] > mid]
return qsort(low)+[mid]+qsort(up)
print(qsort(a))
```
1. 选择基准值
2. 分别找出比基准大的值和比基准小的值
3. 递归的对子数组进行排序
##### 归纳证明
1. 基线条件
2. 归纳条件 证明排序len=1的数组有用 len=2 有用 len=3有用
所以之后无需证明都有效
### 大O表示法
结论:
* 指明的是算法的增速
* 指出的是算法的最糟运行时间
* 不考虑`+*-/`
* 通常不考虑常量(快速查找常量比归并小)
### qsort 时间复杂度
* 最糟糕(头)
1. 以头为基准,需要涉及整个列表
2. 因为要遍历两边的数组,而且O(n)不受常量影响
* 最优(中间)
1. 以中间为基准(类似二分log n)
2. 每层O(n)
最佳的就是平均的情况
TODO: 随机的选择用作基准的元素
```python
array = [5,7,2,4,3,1,8,6]
from random import randint
def qsort(array):
if len(array)<=1:
return array
ran=randint(0,len(array)-1)
mid = array[ran]
array.pop(ran)
smaller = [ i for i in array if i<=mid]
bigger = [i for i in array if i>mid ]
return qsort(smaller) + [mid] +qsort(bigger)
print(qsort(array))
```
### 散列表
DNS 解析
域名->ip地址
### 广度优先遍历
```python
from collections import deque
graph = {}
graph['you'] = ["alice","bob","claire"]
graph['bob'] = ["anuj","peggy"]
graph["alice"] = ["peggy"]
graph["claire"] = ["thom","jooy"]
graph["anuj"] = []
graph["peggy"] = []
graph["thom"] = []
graph["jonny"] = []
def person_is_seller(name):
return name[-1] == 'm'
# deque
def find():
search_queue = deque()
search_queue += graph["you"]
while search_queue:
person = search_queue.popleft()
if person_is_seller(person):
print(person +"is seller")
return True
else:
search_queue +=graph[person]
return False
find()
```
使用一个队列 从右侧加入一个节点的所有的下个节点
然后从左侧读取一个节点 往返直到队列为空
```python
from collections import deque
def person_is_seller(name):
return name[-1] == 'm'
graph = {}
graph['you'] = ["alice","bob","claire"]
graph['bob'] = ["anuj","peggy"]
graph["alice"] = ["peggy"]
graph["claire"] = ["thom","jooy"]
graph["anuj"] = []
graph["peggy"] = []
graph["thom"] = []
graph["jonny"] = []
search_queue = deque()
def find2(search_queue):
if search_queue:
person = search_queue.popleft()
if person_is_seller(person):
print(person +"is seller")
return True
else:
search_queue +=graph[person]
return find2(search_queue)
else:
return None
search_queue += graph["you"]
print(find2(search_queue))
```
* 广度优先遍历是找到是否A-B的
* 有就可以找到最短路径
* 有向图可以指定方向
* 无向图关系双向
* 按照顺序放入队列就可以找到最短路径,检查过的人需要放入去重列表
### 迪杰斯特拉算法
```python
graph = {}
graph['start'] = {}
graph['start']['a'] = 5
graph['start']['b'] = 0
graph['a'] = {}
graph['a']['c'] = 15
graph['a']['d'] = 20
graph['b'] = {}
graph['b']['c'] = 30
graph['b']['d'] = 25
graph['c'] = {}
graph['c']['fin'] = 20
graph['d'] = {}
graph['d']['fin'] = 10
graph['fin'] = {}
costs = {}
costs['a'] = 5
costs['b'] = 0
costs['c'] = float("inf")
costs['d'] = float("inf")
costs['fin'] = float("inf")
parents = {}
parents['a'] = 'start'
parents['b'] = 'start'
parents['c'] = 'start'
parents['d'] = 'start'
parents['fin'] = None
processed = []
def find_lowst(costs):
low_cost = float("inf")
low_cost_node = None
for node in costs:
cost = costs[node]
if cost < low_cost and node not in processed:
low_cost = cost
low_cost_node = node
return low_cost_node
node = find_lowst(costs)
while node is not None:
cost = costs[node]
friends = graph[node]
for friend in friends.keys():
new_cost = friends[friend]+cost
# if new_cost <
if new_cost < costs[friend]:
costs[friend] = new_cost
parents[friend] = node
processed.append(node)
node = find_lowst(costs)
print(costs)
```
简洁版本
```python
graph = {}
graph['start'] = {}
graph['start']['a'] = 5
graph['start']['b'] = 0
graph['a'] = {}
graph['a']['c'] = 15
graph['a']['d'] = 20
graph['b'] = {}
graph['b']['c'] = 30
graph['b']['d'] = 25
graph['c'] = {}
graph['c']['fin'] = 20
graph['d'] = {}
graph['d']['fin'] = 10
graph['fin'] = {}
def get_costs_parent(graph):
costs = {}
parents = {}
for node in graph.keys():
if node in graph['start'].keys():
costs[node] = graph['start'][node]
parents[node] = 'start'
else:
costs[node] = float('inf')
parents[node] = None
return costs,parents
costs ,parents= get_costs_parent(graph)
processed = []
def find_lowst(costs):
low_cost = float("inf")
low_cost_node = None
for node in costs:
cost = costs[node]
if cost < low_cost and node not in processed:
low_cost = cost
low_cost_node = node
return low_cost_node
node = find_lowst(costs)
while node is not None:
cost = costs[node]
friends = graph[node]
for friend in friends.keys():
new_cost = friends[friend]+cost
# if new_cost <
if new_cost < costs[friend]:
costs[friend] = new_cost
parents[friend] = node
processed.append(node)
node = find_lowst(costs)
print(costs['fin'])
```
```python
graph = {}
graph['start'] = {}
graph['start']['a'] = 5
graph['start']['b'] = 0
graph['a'] = {}
graph['a']['c'] = 15
graph['a']['d'] = 20
graph['b'] = {}
graph['b']['c'] = 30
graph['b']['d'] = 25
graph['c'] = {}
graph['c']['fin'] = 20
graph['d'] = {}
graph['d']['fin'] = 10
graph['fin'] = {}
def get_costs_parent(graph):
costs = {}
parents = {}
for node in graph.keys():
if node in graph['start'].keys():
costs[node] = graph['start'][node]
parents[node] = 'start'
else:
costs[node] = float('inf')
parents[node] = None
return costs,parents
costs ,parents= get_costs_parent(graph)
def find_lowst(costs):
low_cost = float("inf")
low_cost_node = None
for node in costs:
cost = costs[node]
if cost < low_cost and node not in processed:
low_cost = cost
low_cost_node = node
return low_cost_node
def find_short_path(costs,processed,parents):
node = find_lowst(costs)
if node is not None:
cost = costs[node]
neighbors = graph[node]
for n in neighbors.keys():
new_cost = neighbors[n] + cost
if new_cost < costs[n]:
costs[n] = new_cost
parents[n] = node
processed.append(node)
return find_short_path(costs,processed,parents)
else:
return costs['fin']
processed = []
fastst = find_short_path(costs,processed,parents)
```
递归实现:
* 基准条件:找到的最小节点为空
* 递归条件:还有节点没有处理
#### 主要思路:
1. 找到一个节点然后取找他的所有相邻节点
2. 将相邻节点到本节点的距离与本节点到起点的距离相加
3. 判断是否比相邻结点原来的距离短
4. 如果更短直接更新,并且加入去重列表
5. 如何取找一个节点:我们选取的是最小的节点,如果这个节点不在去重队列
并且是最小的,就以他为节点更新他的相邻节点,至于我们要选择最小的,是因为
要是选择最大的化会遇到无限大的情况(没有找到到达线路)
* 广度优先搜索可用来在非加权图中查找最短路径
* Dijkstra适合在加权图中查找最短路径
* 加权图为正:Dijkstra/加权图为负:贝尔曼-福德
### 贪心算法
```python
array = ['mt','wa','or','id','nv','ut','ca','az']
states_needed = set(array)
# 转换为集合
stations = {}
stations['kone'] = set(['id','nv','ut'])
stations['ktwo'] = set(['wa','id','mt'])
stations['kthree'] = set(['or','nv','ca'])
stations['kfour'] = set(['nv','ut'])
stations['kfive'] = set(['ca','az'])
stations_array = []
while states_needed:
bast_station = None
states_covered = set()
for station,state_for_station in stations.items():
# print(station,state_for_station)
covered = state_for_station & states_needed
if len(covered) > len(states_covered):
bast_station = station
states_covered = covered
states_needed = states_needed - states_covered
# print(states_needed)
stations_array.append(bast_station) #最好的station
print(stations_array)
```
* 贪心算法思路:每步都选择最有解,从而达到整体最优解
* 不适用场景:背包问题/只能找到非常接近最优解的解法
```python
#递归实现
def find_station(states_needed):
if states_needed:
bast_station = None
states_covered = set()
for station,state_for_station in stations.items():
# print(station,state_for_station)
covered = state_for_station & states_needed
if len(covered) > len(states_covered):
bast_station = station
states_covered = covered
states_needed = states_needed - states_covered
return [bast_station] +find_station(states_needed)
else:
return []
print(find_station(states_needed))
```
实现思路:
1.寻找覆盖未被覆盖区域最多的电台
2.将这个加入队列
3.更新未被覆盖区域
4.重复1-3
### NP完全问题
1. 元素较少运行速度快,元素越来越多速度会变得非常慢.
2. 涉及所有组合的通常是NP完全问题
3. 不能分割成小问题,需要考虑各种情况的
4. 问题涉及旅行商序列等的且难以解决的
5. 问题涉及广播台集合的
6. 问题可以转换为旅行商或者集合覆盖的问题的
```python
graph = {}
graph['a'] = {}
graph['a']['b'] = 3
graph['a']['c'] = 6
graph['a']['d'] = 1
graph['c'] = {}
graph['c']['a'] = 6
graph['c']['b'] = 2
graph['c']['d'] = 7
graph['b'] = {}
graph['b']['c'] = 2
graph['b']['d'] = 5
graph['b']['a'] = 3
graph['d'] = {}
graph['d']['c'] = 7
graph['d']['a'] = 1
graph['d']['b'] = 5
list_city_array=list(graph.keys())
def find_fast(graph,next_array):
if next_array:
finded_array = []
city = next_array.pop()
list_array = list(graph.keys())
cost = 0
finded_array.append(city)
while len(finded_array) < len(list_array):
low_cost = float('inf')
low_cost_city = None
for neibo in graph[city].keys():
if graph[city][neibo]