Question

I have a required where i need to extract a number from a string in the needed format like :

• f001–>100
• f100–>1
• 2030–>302
• 0203–>3020
• 2031–>1302

so the operation above is :

• remove any f character if it is present
• reverse the string
• remove leading zeros from the string

I have written a code which is working fine in c++:

int main(int argc,char* argv[])
{

string str1(argv[argc-1]);


reverse(str1.begin(),str1.end());
str1.erase(remove(str1.begin(),str1.end(),'f'),str1.end());
str1.erase(0,str1.find_first_not_of('0',0));

cout <<str1<<endl;

return 0;

}

is there any better way of doing the same thing?

Answer 1

I Guess, a simple function as below will do the job

NOTE:- It is not a complete program. Just a flow… It will parse the string once in stead of 3 times as in your case. I definitely hope there are better C++ style approach, and looking forward to the same.

foo(char* str) {

int state=0;

int len=strlen(str);

for(i=len-1; i>=0; i++) {

if(state==0 && str[i]!='0') { //Ignore trailing zeros
    state=1;
}
else if(stare==1) {
    if(str[i]=='f')
        break;
    //Reverse logic here, just store the char in a heap as it comes which will be returned after the for finishes
}  

}

}