python基础-递归、二分法查找（for\递归）、三级菜单、压栈思想

递归方法

# age(1) n = 1 age(2)+2
# age(2) n = 2 age(3)+2
# age(3) n = 3 age(4)+2
# age(4) n = 4 40

def age(n):
    if n == 4:
        return 40
    return age(n+1)+2

print(age(1))

输出如下：

E:\python\python_sdk\python.exe E:/python/py_pro/1.初识递归.py
46

Process finished with exit code 0

二分法查找

通过for循环实现

思路是获取头尾index,取中间索引值，与要查找的值进行判断，如果中间值大于要查找的值，那么就在列表的前半部分查找，如果中间值小于要查找的值，那么就在列表后半部分查找，折半查找时候，在更新要查找的切片的头尾索引值

def method(mLis,mVal):
   minIndex = 0
   maxIndex = len(mLis) -1
   while True:
      centerIndex = (minIndex + maxIndex) // 2
      centerValue = mLis[centerIndex]

      if centerValue < mVal:#右边
          minIndex = centerIndex + 1
      if centerValue > mVal:#左边
          maxIndex = centerIndex  - 1
      if centerValue == mVal:
          return centerIndex

print(method(l,18))

输出如下：6

通过递归实现

def find_2(l,aim,start=0,end=None):  #[2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]
    if end == None:end = len(l) - 1  #start = 0,end = 24
    if start <= end:
        mid = (end-start) // 2  + start  #mid = 12
        if l[mid] > aim:
            ret = find_2(l,aim,start,mid-1)
            return ret
        elif l[mid] < aim:              #
            ret = find_2(l,aim,mid+1,end)     #find_2(l,58,13,24)
            return ret
        else:
            return aim,mid
    else:
        print('找不到这个值')
l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]
print(find_2(l,22))

输出如下：

(22, 7)

递归应用–三级菜单

menu = {
    '北京': {
        '海淀': {
            '五道口': {
                'soho': {},
                '网易': {},
            }
        },
        '昌平': {
            '沙河': {
                '北航': {},
            },
            '天通苑': {}
        }
    },
    '上海': {
        '闵行': {
            "人民广场": {
                '炸鸡店': {}
            }
        }
    }
}

def menu_3(menu):
    while True:
        for key in menu:
            print(key)      #北京上海山东
        choice = input('选择 : ')   #北京
        if choice == 'q' or choice == 'b':
            return choice
        elif choice in menu and menu[choice]:        #北京 in menu
            borq =  menu_3(menu[choice])
            if borq == 'q':
                return 'q'

menu_3(menu)

测试如下：

E:\python\python_sdk\python.exe E:/python/py_pro/4.三级菜单.py
北京
上海
选择 : 北京
海淀
昌平
选择 : 海淀
五道口
选择 : 五道口
soho
网易
选择 : b
五道口
选择 : b
海淀
昌平
选择 : b
北京
上海
选择 : b

Process finished with exit code 0

压栈

从该数据结构中返回由指定的字段和对应的值组成的字典，如果指定字段不存在就跳过该字段

这道题使用了压栈的做法

通过data = l.pop()取出最后一个栈内元素赋值变量，意味着将l清空了，如果data里面的键在列表field_lst里面，就添加进一个新的result 字典里面，如果data的键对应的值是字典，那么就添加进来到l列表，例如{ “fld1”:1,”fld2”:2}
循环完毕后，列表里面就包含了[{‘fld1’: 1, ‘fld2’: 2}, {‘fld3’: 0, ‘fld5’: 0.4}, {‘key’: {‘fld19’: 1}}]

然后开始一个个pop获取键值对

fields = "fld2|fld5|fld6|fld19|fld7|fld46"
data={"time":"2016-08-05T13:13:05",
    "some_id":"ID1234",
    "grp1":{ "fld1":1,"fld2":2},
    "xxx2":{ "fld3":0,"fld5":0.4},
    "fld6":{"key":{ "fld19":1}},
    "fld7":7,
    "fld46":8}
#计算机思想:压栈，Alex
def select(data,fields):
    l = [data]
    field_lst = fields.split('|')
    result = {}
    while l:

        data = l.pop()     #data = { "fld1":1,"fld2":2}

        for key in data:
            if type(data[key]) == dict:
                l.append(data[key])   # l = [{ "fld1":1,"fld2":2},{ "fld3":0,"fld5":0.4},{"key":{ "fld19":1}}]
            elif key in field_lst:
                result[key] = data[key]
        print("-----",l)
    return result

print(select(data,fields))

输出如下：

E:\python\python_sdk\python.exe E:/python/py_pro/python.py
----- [{'fld1': 1, 'fld2': 2}, {'fld3': 0, 'fld5': 0.4}, {'key': {'fld19': 1}}]
----- [{'fld1': 1, 'fld2': 2}, {'fld3': 0, 'fld5': 0.4}, {'fld19': 1}]
----- [{'fld1': 1, 'fld2': 2}, {'fld3': 0, 'fld5': 0.4}]
----- [{'fld1': 1, 'fld2': 2}]
----- []
{'fld7': 7, 'fld46': 8, 'fld19': 1, 'fld5': 0.4, 'fld2': 2}

Process finished with exit code 0