#include<iostream>#include<stdlib.h>using namespace std;char *
remove_spaces(char *s){
char *result;
char *temp;
temp=s;
result=s;
while(*s!='\0')
{
if(*s==' ')
{
s++;
}
else
{
*temp = *s;
temp++;
s++;
}
}
*temp = '\0';
return result;}int main(){
char s[1000];
gets(s);
cout<<"string after removing spaces " << remove_spaces(s) << endl ;
}
On Fri, Oct 14, 2011 at 1:10 AM, janani thiru <[email protected]>wrote:
> This solution doesn't work.
>
> It prints a blank string.
>
> I think because sometimes the condition becomes true only when a space is
> encountered. After which when we shift the value what happens to the value
> at the position where we are shifting from. It will still contain the char.
> This would be wrong.
>
>
> Janani T
>
>
>
> On Thu, Oct 13, 2011 at 22:02, Don <[email protected]> wrote:
>
>> @pradad
>>
>> 1. Never put strlen in the condition of a loop. That instantly makes
>> an O(n) loop into O(n^2).
>> 2. Be sure to put the null terminator at the end of the compacted
>> string. Your current code stops one short of doing that.
>> 3. Put {} around the body of the for loop if the body contains more
>> than one semicolon.
>> 4. If there is common code in an if block and the else block, you can
>> move it out of the if/else.
>> 5. The variable i is extraneous.
>>
>> The cleanest way I see to write the code is like this:
>>
>> for(ptr1 = ptr2 = string; *ptr2; ++ptr2)
>> if (*ptr2 != ' ') *(ptr1++) = *ptr2;
>> *ptr1 = 0;
>>
>> Don
>>
>> On Oct 11, 3:41 pm, prasad jondhale <[email protected]> wrote:
>> > *ptr1=*ptr2=string;
>> > for(i=0;i<strlen(string);i++)
>> > if(*str==' ')
>> > ptr2++;
>> > else
>> > {
>> > *ptr1=*ptr2;
>> > ptr1++;
>> > ptr2++;
>> > }
>> >
>> > hey guys if anything wrong in this code pls let me know............
>> >
>> > On Tue, Oct 11, 2011 at 11:04 AM, DIPANKAR DUTTA <
>> [email protected]
>> >
>> >
>> >
>> > > wrote:
>> > > Solution 1:
>> > > need two scan :
>> >
>> > > start from left and replace one by one by non-space character ( thus
>> > > have minimal replacemnt)
>> >
>> > > count =0;
>> > > for (int i=0;i<len(str);i++)
>> > > {
>> > > if(str[i] !=NONSPCECHAR)
>> > > { str[count++]=str[i]
>> > > }
>> >
>> > > }
>> >
>> > > ps: any way it needs Left shift...and all solution must need left
>> > > shift if you try to compact it in lower index area.
>> >
>> > > On 10/11/11, abhishek sharma <[email protected]> wrote:
>> > > > Can in place compaction be done without left shifts?
>> >
>> > > > --
>> > > > Nice Day
>> >
>> > > > Abhishek Sharma
>> > > > Bachelor of Technology
>> > > > IIT Kanpur (2009)
>> >
>> > > > --
>> > > > You received this message because you are subscribed to the Google
>> Groups
>> > > > "Algorithm Geeks" group.
>> > > > To post to this group, send email to [email protected].
>> > > > To unsubscribe from this group, send email to
>> > > > [email protected].
>> > > > For more options, visit this group at
>> > > >http://groups.google.com/group/algogeeks?hl=en.
>> >
>> > > --
>> > > Thanks and Regards,
>> > > ------------------------------
>> > > **DIPANKAR DUTTA
>> > > Software Development Engineer
>> > > Xen Server - OpenStack Development Team (DataCenter and Cloud)
>> >
>> > > Citrix R&D India Pvt Ltd
>> > > 69/3, Millers Road, Bangalore – 560052
>> > > Phone: +91 8147830733
>> > > Office: Extn: 16429
>> > > Email: [email protected]
>> >
>> > > --
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>> Groups
>> > > "Algorithm Geeks" group.
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>> > > For more options, visit this group at
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>> >
>> > --
>> >
>> > Thanks & Regards
>> > Prasad Y.Jondhale
>> > M.Tech(software engg.),
>> > Delhi College Of Engineering,
>> > Main Bawana road,Delhi-110042
>> > Ph-09540208001
>>
>> --
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>>
>>
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--
Regards ,
P Veera Reddy Devagiri
Senior Under Graduate
Computer Science and Engineering
IIIT Hyderabad
Mobile no-+91-9492024783
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