@Dave :- a minor change
Initially, decrement the end pointer till it points to positive number,
Now end points to the last negative number.
Now,
If current is negative , increment current
If current is positive , swap it with the element at end and decrement
current and end both
If current >= end , then break.
Algo :-
cur = 0;
end = size - 1;
while ( a[end] > 0 && end > 0 ) end - - ;
while ( cur <end ) {
if( a[cur] < 0 ) cur++;
else{
swap( a[cur], a[end] );
end - - ;
} // end of if-else
} // end of while
In the above example :- ( 1, -1, 2 ), current points to 1 (cur=0) and end
points to -1 (end =1 ) after end has been decremented.
Now swap the element at current and end pointers.
Now cur = 0, end =0, break condition is reached. the output is :- ( -1, 1,
2 )
Please check.
Navin Kumar Gupta
Final Year, B.Tech (Hons.)
Computer Science & Engg.
National Institute of Technology,Jamshedpur
Mob - (+91) 8285303045
On Friday, 29 June 2012 22:28:15 UTC+5:30, Dave wrote:
>
> @Navin: Try this with {1,-1,2}. current points to the 1 and end points to
> the 2. Since 1 is positive, the algorithm swaps the 1 and the 2, giving
> {2,-1,1}. Then it decrements current to point outside the array and end to
> point to the -1. How can this be right?
>
> Dave
>
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