Classical algorithm-evaluate the maximum length of the countermeasure string (version 2)

ClassicAlgorithmQuestion -- obtain the maximum length of the countermeasure string (version 2)

Method 1: The idea is quite satisfactory. traverse this string. If two adjacent characters are found to be equal, cyclically judge whether the two adjacent characters are equal,
Keep down the number of characters that meet the conditions. The maximum number is required. (This method is suitable for strings such as Google)
Method 2: The train of thought is the same as the method at the moment. It is suitable for strings such as ggoggle.
Method 3: Meet the meaning of the question, suitable for any type of string. The time complexity is O (n ^ 2 ).

Method 1

 Int  Counterplan1 (  Const  String  Str)
{
  Int  Strlen  =  Str. Length ();
 Int  Maxlen  =  0  ;
  For  (  Int  I  =  0  ; I  <  Strlen  -  1  ; I  ++ )
{
  If  (STR [I]  =  STR [I  +  1  ])
{
  Int  Start  =  I  -  1  ;
  Int  End =  I  +  2  ;
  While  (Start  >  0  &&  End  <  Strlen)
{
  If  (STR [start]  =  STR [end])
{
 --  Start;  ++  End;
}  Else  
{
  Break  ;
}
}
  If  (Maxlen  <  End  -  Start  -  1 )
{
Maxlen  =  End  -  Start  -  1  ;
}
}
}
  Return  Maxlen;
} 

Method 2

 Int  Counterplan2 (  Const  String Str)
{
  Int  Strlen  =  Str. Length ();
  Int  Maxlen  =  0  ;
  For  (  Int  I  =  1  ; I <  Strlen  -  1  ; I  ++  )
{
  If  (STR [I  -  1  ]  =  STR [I  +  1  ])
{
 Int  Start  =  I  -  1  ;
  Int  End  =  I  +  2  ;
  While  (Start  >  0 &&  End  <  Strlen)
{
  If  (STR [start]  =  STR [end])
{
  --  Start;  ++  End;
}  Else  
{
  Break ;
}
}
  If  (Maxlen  <  End  -  Start)
{
Maxlen  =  End  -  Start;
}
}

}
  Return  Maxlen;
} 

Method 3

Int  Counterplan3 (  Const  String  Str)
{
  Int  Strlen  =  Str. Length ();
  Int  Maxlen  =  0  ;
  Int  Start  = 0  , End  =  Strlen  -  1  ;
  While  (Start  <  End)
{
  If  (STR [start]  ! =  STR [end])
{
  If  (Start =  End)
{
  ++  Start;
End  =  Strlen  -  1  ;
}  Else  {
  --  End;
  Continue  ;
}
}

  Int I  =  Start, J  =  End;
  Bool  Isok  =  False  ;
  While  (I  <  J  &&  !  Isok ){
  While (STR [  ++  I]  =  STR [  --  J])
{
  If  (I  <  J  -  2  )
{
  Continue  ;
}

  If (End  -  Start  +  1  >  Maxlen)
{
Maxlen  =  End  -  Start  +  1  ;
}
Isok  =  True ;
  Break  ;
}
I  = ++  Start; j  =  End;
}

  If  (Strlen  -  Start  <=  Maxlen  -  1  )
{
 Break  ;
}
}
  Return  Maxlen;
}