Suppose the vector is vec then , vec.sort(vec.begin(),vec.end()); vec.resize(distance(vec.begin(),unique(vec.begin(),vec.end())));
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 166 |
2 | maomao90 | 163 |
2 | Um_nik | 163 |
4 | atcoder_official | 161 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | nor | 153 |
9 | Dominater069 | 153 |
Suppose the vector is vec then , vec.sort(vec.begin(),vec.end()); vec.resize(distance(vec.begin(),unique(vec.begin(),vec.end())));
Name |
---|
i don't know if this will solve ur problem, but maybe u can try using
std::set
instead ofstd::vector
.And what's wrong with your solution? I usually use one of these two snippets:
Isn't your first snippet wrong?
It should be something like this (you've also missed brackets after
end
).why we subtract a.begin() ?
This function removes all but the first element from every consecutive group of equivalent elements in the range [first,last). It returns an iterator to the element that follows the last element not removed. The range between first and this iterator includes all the elements in the sequence that were not considered duplicates.