blocks|key|47714|text|微软/亚马逊求职面试类型的问题:这是伪代码，实际代码留作练习。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|47715|for+each+char+in+the+string+do:
+++if+the+current+char+is+equal+to+the+next+char:
++++++delete+next+char
+++else
+++++continue

return+string|code-block|syntax|javascript|47716|作为更高层次的尝试(而不是实际的实现)：|47717|for+s+in+string:
++if+s+==+s%2B1:++##+check+until+the+end+of+the+string
+++++delete+s%2B1|47718|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Microsoft / Amazon job interview type of question:
This is the pseudocode, the actual code is left as exercise. 

<pre><code>for each char in the string do:
 if the current char is equal to the next char:
 delete next char
 else
 continue

return string
</code></pre>

As a more high level, try (not actually the implementation):

<pre><code>for s in string:
 if s == s+1: ## check until the end of the string
 delete s+1
</code></pre>

blocks|key|46667|text|%2B1表示groupby。即兴的，像这样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|46668|from+itertools+import+groupby
def+remove_dupes(arg):
++++#+create+generator+of+distinct+characters,+ignore+grouper+objects
++++unique+=+(i[0]+for+i+in+groupby(arg))
++++return+''.join(unique)|code-block|syntax|javascript|46669|在Python+2.7.2中为我做饭|46670|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

+1 for groupby. Off the cuff, something like:

<pre><code>from itertools import groupby
def remove_dupes(arg):
 # create generator of distinct characters, ignore grouper objects
 unique = (i[0] for i in groupby(arg))
 return ''.join(unique)
</code></pre>

Cooks for me in Python 2.7.2

blocks|key|2734806|text|首先，您不能从Python中的字符串中删除任何内容(google+"Python不可变字符串“，如果不清楚的话)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2734807|M第一种方法是：|2734808|foo+=+'12233322155552'
bar+=+''
for+chr+in+foo:
++++if+bar+==+''+or+chr+!=+bar[len(bar)-1]:
++++++++bar+%2B=+chr|code-block|syntax|javascript|2734809|或者，使用上面的itertools提示：|2734810|''.join([+k[0]+for+k+in+groupby(a)+])|2734811|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

First of all, you can't remove anything from a string in Python (google "Python immutable string" if this is not clear).

M first approach would be:

<pre><code>foo = '12233322155552'
bar = ''
for chr in foo:
 if bar == '' or chr != bar[len(bar)-1]:
 bar += chr
</code></pre>

or, using the itertools hint from above:

<pre><code>''.join([ k[0] for k in groupby(a) ])
</code></pre>

blocks|key|2734816|text|t+=+'12233322155552'
for+i+in+t:
++++dup+=+i%2Bi
++++t+=+re.sub(dup,+i,+t)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2734817|您可以获得1232152格式的最终输出|unstyled|offset|length|style|CODE|2734818|entityMap^0|0|5|7|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>t = '12233322155552'
for i in t:
 dup = i+i
 t = re.sub(dup, i, t)
</code></pre>

You can get final output as <code>1232152</code>

blocks|key|47770|text|import+re

#+Only+repeated+numbers
answer+=+re.sub(r'(\d)\1%2B',+r'\1',+'12233322155552')

#+Any+repeated+character
answer+=+re.sub(r'(.)\1%2B',+r'\1',+'12233322155552')|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|47771|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>import re

# Only repeated numbers
answer = re.sub(r'(\d)\1+', r'\1', '12233322155552')

# Any repeated character
answer = re.sub(r'(.)\1+', r'\1', '12233322155552')
</code></pre>

blocks|key|47097|text|number+=+'12233322155552'
temp_list+=+[]


for+item+in+number:+++
+++if+len(temp_list)+==+0:
++++++temp_list.append(item)

+++elif+len(temp_list)+>+0:
++++++if++temp_list[-1]+!=+item:
++++++++++temp_list.append(item)

print(''.join(temp_list))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|47098|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>number = '12233322155552'
temp_list = []


for item in number: 
 if len(temp_list) == 0:
 temp_list.append(item)

 elif len(temp_list) &gt; 0:
 if temp_list[-1] != item:
 temp_list.append(item)

print(''.join(temp_list))
</code></pre>

blocks|key|2525378|text|你可以使用itertools，这是一个内联代码|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2525379|>>>+s+=+'12233322155552'
>>>+''.join(i+for+i,+_+in+itertools.groupby(s))
'1232152'|code-block|syntax|javascript|2525380|entityMap^0|5|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can use <code>itertools</code>, here is the one liner 

<pre><code>&gt;&gt;&gt; s = '12233322155552'
&gt;&gt;&gt; ''.join(i for i, _ in itertools.groupby(s))
'1232152'
</code></pre>

blocks|key|47776|text|这将是一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|47777|def+fix(a):
++++list+=+[]

++++for+element+in+a:
++++++++#+fill+the+list+if+the+list+is+empty
++++++++if+len(list)+==+0:list.append(element)
++++++++#+check+with+the+last+element+of+the+list
++++++++if+list[-1]+!=+element:++list.append(element)

++++print(''.join(list))++++


a=+'GGGGiiiiniiiGinnaaaaaProtijayi'
fix(a)
#+output+=>+GiniGinaProtijayi|code-block|syntax|javascript|47778|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This would be a way:

<pre><code>def fix(a):
 list = []

 for element in a:
 # fill the list if the list is empty
 if len(list) == 0:list.append(element)
 # check with the last element of the list
 if list[-1] != element: list.append(element)

 print(''.join(list)) 


a= 'GGGGiiiiniiiGinnaaaaaProtijayi'
fix(a)
# output =&gt; GiniGinaProtijayi
</code></pre>

For a string such as <code>'12233322155552'</code>, by removing the duplicates, I can get <code>'1235'</code>.

But what I want to keep is <code>'1232152'</code>, only removing the consecutive duplicates.

How to remove duplicates only if consecutive in a string?

Python

对于像'12233322155552'这样的字符串，通过删除重复项，我可以得到'1235'。但我想保留的是'1232152'，只删除连续的重复项。

问如何仅当字符串中连续时才删除重复项？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何仅当字符串中连续时才删除重复项？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何仅当字符串中连续时才删除重复项？
EN