Regular expression to remove quotations

Question

How do I write a regular expression to satisfy these requirements ? I can only use a string.replaceAll function ..

a) For ” which appears at end of paragraph which has a “, but not “ “—remove ”

b) For “ which appears at beginning of paragraph remove “ [NOTE: If there is “ “, it should now be “]

c) For ” which appears at end of paragraph without a matching “ at beginning of paragraph –remove ”

EDIT:

Rule a)
Transform:
String input1 ="“remove quotes”" 
String output1 ="“remove quotes"

Don't change anything:
String input1 ="““remove quotes”" 
String output1 ="““remove quotes”"

Rule b)
Transform:
String input1 ="“remove quotes”" 
String output1 ="remove quotes”"

Replace with single ldquo:
String input1 ="““remove quotes”" 
String output1 ="“remove quotes”"

Rule c)
Do nothing (there is a matching ldquo):
String input1 ="“do not remove quotes”" 
String output1 ="“do not remove quotes”"

Transform(no matching ldquo hence remove rdquo):
String input1 ="remove quotes”" 
String output1 ="remove quotes"

I think I am going to run all the 3 rules separately on the string. What would be 3 regexes and replace expressions ? 

Answer 1

Description

This regex will do the following:

1. if 2 initial “ strings and a ending ”, then remove single “
2. if 1 initial “ string and a ending ”, then remove nothing
3. if 0 initial “ strings and a ending ”, then remove ending ”

regex: ^(?=.*?”)“\s*(“)|^(?=.*?”)(“.*?”)|^(?!“)(.*?)”

replace with: $1$2$3

Input text

“ DO NOTHING  ”
“ “ REMOVE INITIAL LD  ”
REMOVE RD  ”

Output text respecitivly

“ DO NOTHING  ”
“ REMOVE INITIAL LD ”
REMOVE RD

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These expressions where hashed out from a chat session, and written to be executed one at a time in A,B,C order, however because they are seperate, they can be executed in any order the developer would like which would change based on the desired output.

A

• 1 LD and 1 RD, remove the RD
• 2 LD and 1 RD, do nothing
• regex: ^(“(?!\s*“).*?)”
• replace with $1

B

• 1 LD, remove 1 LD
• 2 LD, remove 1 LD
• regex: ^“(\s*(?:“)?)
• replace with $1

C

• 1 LD and 1 RD, do nothing
• 0 LD and 1 RD, remove the RD
• regex: ^(?!“)(.*?)”
• replace with $1