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[已解决]小甲魚課後練習題

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TsaiYen
发表于 2020-5-9 01:51:18 | 显示全部楼层 |阅读模式

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各位魚友悶~這題可以用切片解出來嗎? 我自己寫的code在最下面!!!!!!
找出小甲鱼藏在下边这个长字符串中的密码，密码的埋藏点符合以下规律
    a. 每位密码为单个小写字母
    b. 每位密码的左右两边均有且只有三个大写字母
str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
-------------------------------------------------------------------------------------------------------------------------
str2 = 'a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z'
str3 = 'A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z'
list1 = []
for each in str1:
    if each not in list1:
        if each in str2:
            idx = str1.index(each)
            a,b,c = str1[idx - 3 : idx]
            a, b, c in str3
            if each in str2:
                idx = str1.index(each)
                d,e,f = str1[idx + 1 : idx + 4]
                d, e, f in str3
                list1.append(each)
                print(each)
最佳答案
月排行榜 / 总排行榜
txxcat
2020-5-9 18:20:34
你更新的代码还是没有解决index只能得到第一个字母位置的问题，然后if each not in list1完全没必要，没有说密码不能有重复的字母，最后是 你判断前三后三是大写，却没有判断前三之前还后三之后那个是不是大写，这也会导致问题。根据你的思路改写了一下代码，可以跑出来，不过有一点不严谨的地方，看看你能找出来不。
还有个建议，判断大小写用isupper()和islow()更好。
list1 = []
lenth=len(str1)            #获取字符串长度
for idx in range(3,lenth-3):                #从第四个开始算到倒数第三个字符
    if str1[idx] in str2 and str1[idx-4] not in str3 and str1[idx+4] not in str3:     #除了判断是小写，加入判断前4后4的字符是否是大写的判断
        a,b,c = str1[idx - 3 : idx]
        d,e,f = str1[idx + 1 : idx + 4]
        if a in str3 and b in str3 and c in str3 and d in str3 and e in str3 and f in str3:
            list1.append(str1[idx])
print(''.join(list1))           #优化一下输出，其实不用列表，用字符相加更方便      
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_2_
发表于 2020-5-9 09:23:04 | 显示全部楼层
本帖最后由 _2_ 于 2020-5-9 13:54 编辑 
import re
str1="…" #这里自行补充
re.findall(r"(A-Z)\1\1(a-z)\1\1\1", str1)
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MIke_python小小
发表于 2020-5-9 10:26:42 | 显示全部楼层
你這個繁體中文看的我很難受的
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txxcat
发表于 2020-5-9 11:03:36 | 显示全部楼层
说几个显性错误：
str1.index(each)获得索引值不靠谱，因为无论each出现多少次，只会返回each再str1中第一次出现的位置的值；
a, b, c in str3、a, b, c in str3，这两句没有任何意义，我猜你是想要：if a in str3 and b in str3 and c in str3:
str2、str3中的字母用逗号隔开没有必要，相反如果str1中有逗号反而容易误判。
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TsaiYen
 楼主| 发表于 2020-5-9 12:01:50 | 显示全部楼层
txxcat 发表于 2020-5-9 11:03
说几个显性错误：
str1.index(each)获得索引值不靠谱，因为无论each出现多少次，只会返回each再str1中第一 ...

這題用切片解的出來嗎
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txxcat
发表于 2020-5-9 12:06:58 | 显示全部楼层
TsaiYen 发表于 2020-5-9 12:01
這題用切片解的出來嗎

切片是解这题的必要操作，算法就各有不同了。
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TsaiYen
 楼主| 发表于 2020-5-9 12:52:53 | 显示全部楼层
txxcat 发表于 2020-5-9 12:06
切片是解这题的必要操作，算法就各有不同了。

只跑出 l  不知道哪裡出問題!!!!!
str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
lqNsYCyaQXBzrFUbkAUAWAKrDgDtAlGMBqWQhpEwquZqWZJpslUfMllCwWptqINjrOBTLuPzwvXNbLCx
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qKWGc'''

str2 = 'a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z'
str3 = 'A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z'
list1 = []           #創建空列表
for each in str1:    #把字串中的字符給each
    if each not in list1:          #判斷each是不是沒有在列表中(沒有的話)
        if each in str2:           #each 判斷是不是小寫字母
            idx = str1.index(each) #找出他在str1裡面的位移植丟給變數
            a,b,c = str1[idx - 3 : idx]
            d,e,f = str1[idx + 1 : idx + 4] 
            if a in str3 and b in str3 and c in str3 and d in str3 and e in str3 and f in str3:
                list1.append(each)
print(list1)
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TsaiYen
 楼主| 发表于 2020-5-9 13:19:54 | 显示全部楼层 
#修改過後只跑出一個 l
str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''

str2 = 'abcdefghijklmnopqrstuvwxyz'
str3 = 'ABCDEFGHJIKLMNOPQRSTUVWXYZ'
list1 = []           #創建空列表
for each in str1:    #把字串中的字符給each
    if each not in list1:          #判斷each是不是沒有在列表中(沒有的話)
        if each in str2:           #each 判斷是不是小寫字母    
            idx = str1.index(each) #找出他在str1裡面的位移植丟給變數
            a,b,c = str1[idx - 3 : idx]
            d,e,f = str1[idx + 1 : idx + 4] 
            if a in str3 and b in str3 and c in str3 and d in str3 and e in str3 and f in str3:
                list1.append(each)
print(list1)
                        
             
        
        
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txxcat
发表于 2020-5-9 18:20:34 | 显示全部楼层    本楼为最佳答案   
你更新的代码还是没有解决index只能得到第一个字母位置的问题，然后if each not in list1完全没必要，没有说密码不能有重复的字母，最后是 你判断前三后三是大写，却没有判断前三之前还后三之后那个是不是大写，这也会导致问题。根据你的思路改写了一下代码，可以跑出来，不过有一点不严谨的地方，看看你能找出来不。
还有个建议，判断大小写用isupper()和islow()更好。
list1 = []
lenth=len(str1)            #获取字符串长度
for idx in range(3,lenth-3):                #从第四个开始算到倒数第三个字符
    if str1[idx] in str2 and str1[idx-4] not in str3 and str1[idx+4] not in str3:     #除了判断是小写，加入判断前4后4的字符是否是大写的判断
        a,b,c = str1[idx - 3 : idx]
        d,e,f = str1[idx + 1 : idx + 4]
        if a in str3 and b in str3 and c in str3 and d in str3 and e in str3 and f in str3:
            list1.append(str1[idx])
print(''.join(list1))           #优化一下输出，其实不用列表，用字符相加更方便      
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TsaiYen
 楼主| 发表于 2020-5-10 16:49:08 | 显示全部楼层
txxcat 发表于 2020-5-9 18:20
你更新的代码还是没有解决index只能得到第一个字母位置的问题，然后if each not in list1完全没必要，没有 ...

想請問下不用這種思路還能用甚麼方法嗎?
for循環不是把str1一個個丟給each為甚麼還是只能得到第一個字母位置
萬分的感謝您!!!!!!!!!!!!!!!!!!!!!
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txxcat
发表于 2020-5-10 20:51:54 | 显示全部楼层
TsaiYen 发表于 2020-5-10 16:49
想請問下不用這種思路還能用甚麼方法嗎?
for循環不是把str1一個個丟給each為甚麼還是只能得到第一個字母 ...

遍历字符串可以获得每一个字符，但你要获得字符串的位置，使用了str1.index(each)，比如'a'这个字符，第一个a在第31个，获得的值是31，这是对的，但是后面的'a'呢，str1.index(each)还是会返回第一个'a'的位置，不会返回你希望的值。如果你希望遍历时获得索引值，还可以这么写：
idx=0
for each in str1:
    pass
    idx+=1
这样可以获得正确的索引值，str1[idx]等于each，但是因为有索引值-3、+3、-4、+4的操作，要正确处理索引值溢出的问题，最后需要的代码会比我的方法要多不少。
这里有几种算法，其实大同小异：
lenth = len(str1)
for i in range(3,lenth-3):
    allupper=True                                 #定义一个标量标志全部是大写
    if str1[i].islower():
        for each in str1[i-3:i]+str1[i+1:i+4]:    #遍历前3个+后3个字符
            if not each.isupper():                #有1个不是大写就跳出判断，标志改为False
                allupper=False
                break
        if allupper and (str1[i-4].islower()) and (str1[i+4].islower()):
                print(str1[i],end="")
暴力简单去除\n法：
str1=str1.replace('\n','')
length = len(str1)
for i in range(3, length-3):
    if str1[i].islower() and  str1[i-3:i].isupper() and str1[i+1:i+4].isupper() and  not str1[i-4].isupper() and not str1[i+4].isupper():
        print(str1[i],end='')
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TsaiYen
 楼主| 发表于 2020-5-11 02:49:39 | 显示全部楼层
txxcat 发表于 2020-5-10 20:51
遍历字符串可以获得每一个字符，但你要获得字符串的位置，使用了str1.index(each)，比如'a'这个字符，第 ...

OK感恩!!! 還在就學嗎?!好厲害
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txxcat
发表于 2020-5-12 00:03:55 | 显示全部楼层
TsaiYen 发表于 2020-5-11 02:49
OK感恩!!! 還在就學嗎?!好厲害

No，作为孩子她爹，对年轻人们充满了羡慕！
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TsaiYen
 楼主| 发表于 2020-5-12 12:31:38 From FishC Mobile | 显示全部楼层
是軟體工程師嗎哈哈！！
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Pythonnewers
发表于 2020-5-12 12:37:36 | 显示全部楼层
MIke_python小小 发表于 2020-5-9 10:26
你這個繁體中文看的我很難受的

哈哈,台灣人?
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TsaiYen
 楼主| 发表于 2020-5-13 13:30:19 From FishC Mobile | 显示全部楼层
對！
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