>> random.uniform(1,20)
1.0130916166719703

習題1：生成[“z1”,”y2”,”x3”,”w4”,”v5”]

#coding=utf-8
result = []
for i in range(1,6):
    result.append(chr(122-i+1)+str(i))
print(result) 

#coding=utf-8
result = []
for i in range(1,6):
result.append(chr(97+i-1)+str(i))
print(result)

習題2：拼接一個字符串的首字母、末尾字母、中間字母為一個字符串
要考慮奇數偶數長度，如果只有1個2個字母的字符串

方式1：直接拼字符串

s = "abcdefghig"

result_str = ""
result_str += s[0]

if len(s)%2 == 1:
    middle_letter = s[len(s)//2]
else:
    middle_letter = s[len(s)//2-1]
    middle_letter += s[len(s)//2]

print(middle_letter)
result_str += middle_letter

result_str += s[-1]

print("拼接後的字符串: ",result_str)
 

方式2：利用列表

def join_str(s):
    result = []
    result.append(s[0])

    if len(s)%2==1:
        result.append(s[len(s)//2])
    else:
        result.append(s[len(s)//2-1])
        result.append(s[len(s)//2])

    result.append(s[-1])
    return "".join(result)

s = "abcdefghig"
s2 = "abcdefghi"
print("拼接後的字符串: ",join_str(s))
print("拼接後的字符串: ",join_st
 

切片沒有越界

>> s = "a"
>> s[5:] #沒有越界
‘‘
>> s[5]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: string index out of range

>> a = [1]
>> a[0:1] = [1,2,3,4]
>> a
[1, 2, 3, 4]
>> a =[1]
>> a[0:2] = [1,2,3,4]
>> a
[1, 2, 3, 4]

>> a = [1]
>> a[0]
1
>> a[0] = [1,2,3,4]
>> a
[[1, 2, 3, 4]]

>> a = [1]
>> a[0:7]
[1]
>> a[0:7] = [1,2,3,4]
>> a
[1, 2, 3, 4]

>> a
[1, 2, 3, 4]
>> a[0:2] = [7,8,9,10]
>> a
[7, 8, 9, 10, 3, 4]

>> a[0] = [55,66]
>> a
[[55, 66], 8, 9, 10, 3, 4]

>> a[0:5] = [1]
>> a
[1, 4]

習題3：S = “i am, a boy ”把boy替換為m
利用切片賦值特性

算法：先把字符串轉換為列表
遍歷列表，找到boy字符串
替換boy字符串為m

#coding=utf-8
s = "i am, a    boy!!"

list_s = list(s)
sub_length = len("boy")

for i in range(len(list_s)):
    if "".join(list_s[i:i+sub_length]) == "boy":
        list_s[i:i+sub_length] = "m"#可以用切片直接替換
print("".join(list_s)) 

習題4：非遞歸實現生成斐波那契數列

#coding=utf-8
def fib(n):
    result = []
    if n <=0 or not isinstance(n,int):
        return result
    for i in range(n):
        if i < 2:
            result.append(1)
        else:
            result.append(result[i-1] +  result[i-2])
    return result
print("斐波那契數列: ",fib(5))
print("斐波那契數列: ",fib(-2))

習題5：非遞歸實現求第n項斐波那契數列值
算法：遍歷n，如果n等於0或1直接返回1
如果n >=2,記錄前面兩個項的和，並重新記錄新的第1個，第2個數；

#coding=utf-8
def fib(n):

    a,b=1,1
    sum = 0
    if n <=0 or not isinstance(n,int):
        return -1
    for i in range(n):
        if i < 2:
            sum = 1
        else:
            sum = a + b
            a = b
            b = sum
    return sum

print(fib(5))

習題6：遞歸求兩個數的最大公約數
方式1：

算法：

找出兩個數的較小數（第一次）
如果兩個數除以較小數余數都等於0的話，添加到一個列表中
不然的話繼續調用函數自身，並且num-1傳入到num參數中。

def max_common_divisor(a,b,num=0,result=[]):

    if num == 0:
        if a < b:
            num = a
        else:
            num = b

    if a%num == 0 and b%num == 0:
        print(result)
        result.append(num)
    else:

       max_common_divisor(a,b,num-1)

    return result#函數遞歸調用結束後才會執行此代碼

print(max_common_divisor(24,120))    

??????

方式2：更相減損法

def get(small,big):
    if small > big:
        small,big = big,small

    if small == big:
        return small

    return get(small,big-small)

print(get(30,24))

方式3:輾轉相除法

def?gain(small,?big):
????if?small?>?big:
????????small,?big?=?big,?small
????if?small?==?0:
????????return?big
????return?gain(small,?big?%?small)

習題7：利用遞歸，處理嵌套列表，生成列表[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
定義全局變量：（盡量不使用全局變量）

l = [1,2,[3,4,[5,6,7,[8,9,[10,11]]]]]
result  = []
def func(p):
    global result
    for v in p:
        if not isinstance(v,list):
            result.append(v)
        else:
            func(v)
    return result
print(func(l))

引用傳參：

def get_list(l,result=[]):
    for v in l:
        if not isinstance(v,list):
            result.append(v)
        else:
            get_list(v)
    return result

l = [1,2,[3,4,[5,6,7,[8,9,[10,11]]]]]
print(get_list(l))

python學習(13)