@Ankur: +1
On Fri, Oct 7, 2011 at 8:44 PM, Ankur Garg <ankurga...@gmail.com> wrote:
> To do inplace stack reversal u need a mechanism by which u can transfer the
> first node to last of the stack making sure u have count of other nodes
> here
>
> Say u have stack
>
> 5<- top
> 4
> 3
> 2
> 1
>
> The answer should generate the stack as
> 1-<top
> 2
> 3
> 4
> 5
> This give hints to recursion . A normal recursive func willlook like
>
> InplaceReverse(stack<int>&s){
> if(s.empty()) return;
> temp=s.top();
> s.pop();
> InplaceReverse(s);
> }
>
> Now if one had to print the stack nodes from bottom to top this would have
> helped as we get to bottom of stack.But we need a storing mechanism so may
> be one more recursive function can help so that we can pop the nodes first
> and then insert the top node
>
> Say the stack at a pt of time is
> 1<-top
> 2
>
> Then to insert 3 at bottom we shud have some mechanism which can store
> values 1 and 2
>
> So u pop 1 and pop 2 ..Insert 2 and then Insert 1
> 1<- top
> 2
> 3
>
> Only way to achieve this without using extra space is another recursion .So
> I write a helper func which can help me here to achieve this. I call this
> helper Func as something like ReverseHelper and push the temp variable to it
> .
>
> So
>
> InplaceReverse(stack<int>&s){
> if(s.empty()) return;
> temp=s.top();
> s.pop();
> InplaceReverse(s);
> ReverseHelper(s,temp);
> }
> ReverseHelper(stack<int>&s,int val){
> if(s.empty()) {
> s. push(val);
> return;
> }
> temp = s.pop();
> ReverseHelper(s,val);
> s.push(temp);
> }
>
> Remember to pass stack as reference
>
> Regards
> Ankur
>
> }
>
>
>
>
> On Fri, Oct 7, 2011 at 7:54 PM, sravanreddy001
> <sravanreddy...@gmail.com>wrote:
>
>> in place means:
>>
>> use constant extra space
>>
>> in simple terms O(1) space
>>
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