递归和二分法

标签：索引   一个   sdi   def   path   函数   turn   二分   new   

递归(难点)
        自己调用自己
        递归的入口
        递归的出口: return
        递归的循环条件：动
        循环树形结构

# import os
# def func(file_path,ceng):
#     lst = os.listdir(file_path)
#     for file in lst:
#         full_path = os.path.join(file_path, file)
#         if os.path.isdir(full_path):
#             print("\t"*ceng, file)
#             func(full_path, ceng+1)
#         else:
#             print("\t"*ceng, file)
#     else:
#         return
# func("F:\python视频",0)

   二分法查找
        核心思想: 掐头结尾取中间.
        前提条件: 有序.

lst = [1, 8, 16, 32, 55, 78, 89, 1, 5, 4, 1, 3, 4, 8, 9, 5]
# #二分法查找数字首先要排序
lst = sorted(lst)
# print(lst)
#[1, 1, 1, 3, 4, 4, 5, 5, 8, 8, 9, 16, 32, 55, 78, 89]
n = int(input("请输入一个数字: "))
left = 0
right = len(lst)-1
while left <= right:
    mid = (left + right) // 2  #索引只能是整数
    if n > lst[mid]:
        left = mid+1
    elif n < lst[mid]:
        right = mid - 1
    else:
        print("刚刚好,所在位置%s" % mid)
        break
else:
    print("查找的数不存在")
#函数第一套方案
def func(n,lst):
    left = 0
    right = len(lst)-1
    if left <= right:
        mid = (left+right)//2
        if n > lst[mid]:
            new_list = lst[mid+1:]
            return func(n, new_list)#####################
        elif n < lst[mid]:
            new_list =lst[: mid]
            return func(n, new_list)#####################
        else:
            print("刚刚好,所在位置%s"% mid)
            return True
    else:
        return False
print(func(3, lst))
#函数第二套方案
def func(n, lst, left, right):
    if left <= right:
        mid = (left+right)//2
        if n > lst[mid]:
            left = mid + 1
        elif n < lst[mid]:
            right = mid - 1
        else:
            return True
        return func(n, lst, left, right)
    else:
        return False
ret = func(89,lst,0,len(lst)-1)
print(ret)
#关键字参数
def func(n, lst, left=0, right=None):
    if right == None:
        right = len(lst)-1
    if left <= right:
        mid = (left+right)//2
        if n > lst[mid]:
            left = mid + 1
        elif n < lst[mid]:
            right = mid - 1
        else:
            return True
        return func(n, lst, left, right)
    else:
        return False
ret = func(90,lst)
print(ret)

递归和二分法

标签：索引   一个   sdi   def   path   函数   turn   二分   new   

原文地址：https://www.cnblogs.com/verena/p/9683102.html