二分法查找

# 二分法查找主要的作用就是查找元素
# lst = [1,3,5,7,12,36,68,79] # 数据集 百万级数据
# num = int(input("请输入你要查找的元素信息:"))
#
# for el in lst:
#     if num == el:
#         print("存在")
#         break
# else:
#     print("不存在")
# len(lst)
#  0 1 2 3 4  5  6  7
# [1,3,5,7,12,36,68,79]
# n = 28 # 3
# 规则. 掐头结尾取中间, 必须是有序序列,

# 100000000


# 二分法查找 (需要你明白和掌握)
# lst = [1,3,5,7,12,36,68,79]
# n = int(input("请输入一个数"))
#
# left = 0
# right = len(lst) - 1
#
# while left <= right:
#     mid = (left + right)//2
#     if n > lst[mid]:
#         left = mid + 1
#     elif n < lst[mid]:
#         right = mid - 1
#     else:
#         print("存在")
#         break
#
# else:
#     print("不存在")



# 递归
# def func(n, lst):
#
#     left = 0
#     right = len(lst) - 1
#     if lst != []:
#         mid = (left + right)//2
#         if n > lst[mid]:
#             func(n, lst[mid+1:]) # 改变列表
#         elif n < lst[mid]:
#             func(n, lst[:mid])
#         else:
#             print("找到了")
#             return
#     else:
#         print("没找到")
#         return
#
# n = int(input("请输入你要查找的数:"))
# func(n, [1,3,5,7,12,36,68,79]) # 78



# 递归二 (需要你明白和掌握)
# def func(n, lst, left, right): # 递归找到什么是可以变的. 什么是不可以变的
#     if left <= right:
#         mid = (left + right) // 2
#         if n > lst[mid]:
#             left = mid + 1
#             return func(n, lst, left, right)
#         elif n < lst[mid]:
#             right = mid - 1
#             return func(n, lst, left, right) # 递归如果有返回值. 所有调用递归的地方必须写return
#         else:
#             print("找到了")
#             return mid  # 难点
#     else:
#         print("找不到")
#         return -1
#
# n = int(input("请输入你要查找的数:"))
# lst = [1,3,55,98,37,41,2,5,1,4]
# ret = func(n, lst, 0, len(lst)-1) # 78
# print(ret)



#  最快的查找
lst = [1,3,55,98,37,41,2,5,1,4]

new_lst = []
for i in range(99):
    new_lst.append(0)


for i in lst:    # 1,3,55,98
    new_lst[i] = 1

print(new_lst)


i = int(input('请输入你要找的数据'))
if new_lst[i] == 0: # 1次
    print("不存在")
else:
    print("存在")