set<int>s;
s.erase(s.begin());
I wanted to know whether removing the minimum element in set is a O(logN) or O(1) operation ?
I am quite confused about this situation.
Thanks in advance..
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 3947 |
2 | jiangly | 3734 |
3 | Radewoosh | 3646 |
4 | jqdai0815 | 3620 |
4 | Benq | 3620 |
6 | orzdevinwang | 3612 |
7 | ecnerwala | 3581 |
8 | Geothermal | 3569 |
8 | cnnfls_csy | 3569 |
10 | ksun48 | 3479 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | awoo | 162 |
2 | maomao90 | 160 |
3 | nor | 156 |
4 | cry | 155 |
4 | adamant | 155 |
4 | atcoder_official | 155 |
4 | -is-this-fft- | 155 |
8 | maroonrk | 153 |
9 | SecondThread | 147 |
10 | Petr | 146 |
set<int>s;
s.erase(s.begin());
I wanted to know whether removing the minimum element in set is a O(logN) or O(1) operation ?
I am quite confused about this situation.
Thanks in advance..
Название |
---|
It takes amortized constant time. See here set::begin and here set::erase.
Thanks buddy
It says just "constant" (doesn't say about amortisation), but I doubt that's true.
"For the first version (erase(position)), amortized constant." (where position is an iterator) from http://www.cplusplus.com/reference/set/set/erase/
ah, sorry, I just followed Jakube's links, but both of them apparently go to std::set::begin
Yes, I noticed that too :)
Sorry, I fixed the link.