有一道題： 比較兩個(gè)列表范圍，如果包含的話，返回TRUE，否則FALSE。 詳細(xì)題目如下：
Create a function, this function receives two lists as parameters, each list indicates a scope of numbers, the function judges whether list2 is included in list1.
Function signature:
differ_scope(list1, list2)
Parameters:
list1, list2 - list1 and list2 are constructed with strings,
each string indicates a number or a scope of
numbers. The number or scope are randomly, can
be overlapped. All numbers are positive.
E.g.
['23', '44-67', '12', '3', '20-90']
Return Values:
True - if all scopes and numbers indicated by list2 are included in list1.
False - if any scope or number in list2 is out of the range in list1.
Examples:
case1 - list1 = ['23', '44-67', '12', '3', '20-90']
list2 = ['22-34', '33', 45', '60-61']
differ_scope(list1, list2) == True
case2 - list1 = ['23', '44-67', '12', '3', '20-90']
list2 = ['22-34', '33', 45', '60-61', '100']
differ_scope(list1, list2) == False
貼上自己寫的代碼如下:(備注： python 2.7.6)
def differ_scope(list1, list2): print "list1:" + str(list1) print "list2:" + str(list2) #設(shè)置臨時(shí)存放列表 list1_not_ = [] #用于存放列表1正常的數(shù)字值，當(dāng)然要用int()來轉(zhuǎn)換 list1_yes_ = [] #用于存放列表1中范圍值如 44-67 list1_final = [] #用于存放列表1中最終范圍值 如:[1,2,3,4,5,6,7,8,9,10] temp1 = [] list2_not_ = [] #用于存放列表2正常的數(shù)字值，當(dāng)然要用int()來轉(zhuǎn)換 list2_yes_ = [] #用于存放列表2中范圍值如 44-67 list2_final= [] #用于存放列表2中最終范圍值 如:[1,2,3,4,5,6,7,8,9,10] temp2 = [] temp = [] #用于存放列表1,與列表2比較后的列表,從而判斷結(jié)果為True還是False. #對(duì)列表1進(jìn)行處理 for i in range(len(list1)): #用FOR循環(huán)對(duì)列表1進(jìn)行遍歷 tag = 0 if list1[i].find('-')>0:#對(duì)含范圍的數(shù)字進(jìn)行處理，放到list_yes_列表中 strlist = list1[i].split('-') list1_yes_ = range(int(strlist[0]),int(strlist[1])+1)#讓其生成一個(gè)范圍列表 for each in list1_yes_: #FOR循環(huán)遍歷所有符合條件的. [temp1.append(each)] else: #對(duì)列表1中正常的數(shù)字進(jìn)行處理，放到list_not_列表中 list1_not_.append(int(list1[i]))#對(duì)列表1中進(jìn)行處理，放到list_yes_ [temp1.append(i) for i in list1_not_ if not i in temp1]#去除重復(fù)項(xiàng) list1_final = sorted(temp1) #比較后,排序,并放到list1_final列表中 print "list1_final value is:" + str(list1_final)#打印排序后最終list1_final列表 #對(duì)列表2進(jìn)行處理 for i in range(len(list2)): if list2[i].find('-')>0: strlist = list2[i].split('-') list2_yes_ = range(int(strlist[0]),int(strlist[1])+1) for each in list2_yes_: [temp2.append(each)] print "Temp2:" + str(temp2) else: list2_not_.append(int(list2[i])) [temp2.append(i) for i in list2_not_ if not i in temp2] list2_final = sorted(temp2) print "list2_final value is:" + str(list2_final) #對(duì)兩個(gè)列表進(jìn)行比較,得出最終比較結(jié)果. [temp.append(i) for i in list2_final if not i in list1_final]#比較兩個(gè)列表差值. print "In list2 but not in list1:%s" % (temp)#打印出列表1與列表2的差值 if len(temp)>=1 : print "The result is: False" else: print "The result is: True" if __name__ == '__main__': list1 = ['23', '44-67', '12', '3','90-100'] list2 = ['22-34', '33', '45'] differ_scope(list1,list2) 總結(jié)：
1. 這道題關(guān)鍵是想法，如果整成坐標(biāo)的方式來比較，會(huì)很麻煩。
2. 列表轉(zhuǎn)成范圍后，如果消除重復(fù)項(xiàng)，同樣是里面的關(guān)鍵所在。
3. 其次是對(duì)列表遍歷的操作，同樣挺重要。
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