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
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.
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
- 当调用另一个函数时候,当前函数处于未完成状态.
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- 递归指的是调用自己的函数
-
- 递归条件:函数调用自己
- 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
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<y:
y=y%x
if y ==0:
return x
return DC(x,y)分治法思路
- 找出基线条件/尽可能简单
- 不断分割问题->符合基线条件
#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))#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
- 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
- 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
a = [1,2,3,4,5]
def count(array):
if array == []:
return 0
else:
return 1+count(array[1:])
print(count(a))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
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
- 只剩下一个元素,就是那个需要的元素
- 中点直接是猜到的元素
- 递归条件
- 猜的元素比中点大
- 猜的元素比中点小
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))- 选择基准值
- 分别找出比基准大的值和比基准小的值
- 递归的对子数组进行排序
- 基线条件
- 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)
- 最糟糕(头)
- 以头为基准,需要涉及整个列表
- 因为要遍历两边的数组,而且O(n)不受常量影响
- 最优(中间)
- 以中间为基准(类似二分log n)
- 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
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))DNS 解析 域名->ip地址
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
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
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
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'])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
- 找到一个节点然后取找他的所有相邻节点
- 将相邻节点到本节点的距离与本节点到起点的距离相加
- 判断是否比相邻结点原来的距离短
- 如果更短直接更新,并且加入去重列表
- 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
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
#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
- 元素较少运行速度快,元素越来越多速度会变得非常慢.
- 涉及所有组合的通常是NP完全问题
- 不能分割成小问题,需要考虑各种情况的
- 问题涉及旅行商序列等的且难以解决的
- 问题涉及广播台集合的
- 问题可以转换为旅行商或者集合覆盖的问题的
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]<low_cost and neibo not in finded_array:
low_cost = graph[city][neibo]
low_cost_city = neibo
# print(low_cost)
cost += low_cost
finded_array.append(low_cost_city)
city = low_cost_city
mined = min(find_fast(graph,next_array),cost)
return mined
else:
return float("inf")
print(find_fast(graph,list_city_array))旅行商问题使用近似算法实现
- 基准条件:剩余被查找队列为空
- 递归条件:剩余查找队列不为空
1.从城市列表中获得一个城市
2.寻找这个城市最近距离的城市,且不再已查找列表 3.更新时间花费,将被查找城市加入已查找队列
4.以这个城市为起点查找最近距离城市 5.重复2-4
问题:为什么第一排每个单元格都是1500? 回答:第一个单元格指的是第一个1磅时候最大价格是1500 第二个单元格指的是2磅的时候价格是1500 以此类推
我想应该把他们分开