About answer 3 removing only the last \n off string code :
if (!s.empty() && s[s.length()-1] == '\n') {
s.erase(s.length()-1);
}
Will the if condition not fail if the string is really empty ?
Is it not better to do :
if (!s.empty())
{
if (s[s.length()-1] == '\n')
s.erase(s.length()-1);
}
s.erase(std::remove(s.begin(), s.end(), '\n'), s.end());
Use std::algorithms. This question has some suitably reusable suggestions Remove spaces from std::string in C++
Another way to do it in the for loop
void rm_nl(string &s) {
for (int p = s.find("\n"); p != (int) string::npos; p = s.find("\n"))
s.erase(p,1);
}
Usage:
string data = "\naaa\nbbb\nccc\nddd\n";
rm_nl(data);
cout << data; // data = aaabbbcccddd
If its anywhere in the string than you can't do better than O(n).
And the only way is to search for '\n' in the string and erase it.
for(int i=0;i<s.length();i++) if(s[i]=='\n') s.erase(s.begin()+i);
For more newlines than:
int n=0;
for(int i=0;i<s.length();i++){
if(s[i]=='\n'){
n++;//we increase the number of newlines we have found so far
}else{
s[i-n]=s[i];
}
}
s.resize(s.length()-n);//to delete only once the last n elements witch are now newlines
It erases all the newlines once.
You should use the erase-remove idiom, looking for '\n'
. This will work for any standard sequence container; not just string
.
std::string some_str = SOME_VAL;
if ( some_str.size() > 0 && some_str[some_str.length()-1] == '\n' )
some_str.resize( some_str.length()-1 );
or (removes several newlines at the end)
some_str.resize( some_str.find_last_not_of(L"\n")+1 );
If the newline is expected to be at the end of the string, then:
if (!s.empty() && s[s.length()-1] == '\n') {
s.erase(s.length()-1);
}
If the string can contain many newlines anywhere in the string:
std::string::size_type i = 0;
while (i < s.length()) {
i = s.find('\n', i);
if (i == std::string:npos) {
break;
}
s.erase(i);
}
Here is one for DOS or Unix new line:
void chomp( string &s)
{
int pos;
if((pos=s.find('\n')) != string::npos)
s.erase(pos);
}
The code removes all newlines from the string str
.
O(N) implementation best served without comments on SO and with comments in production.
unsigned shift=0;
for (unsigned i=0; i<length(str); ++i){
if (str[i] == '\n') {
++shift;
}else{
str[i-shift] = str[i];
}
}
str.resize(str.length() - shift);
All these answers seem a bit heavy to me.
If you just flat out remove the '\n' and move everything else back a spot, you are liable to have some characters slammed together in a weird-looking way. So why not just do the simple (and most efficient) thing: Replace all '\n's with spaces?
for (int i = 0; i < str.length();i++) {
if (str[i] == '\n') {
str[i] = ' ';
}
}
There may be ways to improve the speed of this at the edges, but it will be way quicker than moving whole chunks of the string around in memory.
Source: Stackoverflow.com