Python 二分法查找

Posted 2021-01-27 缥缈映苍穹

# 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")