Question

见a and b> be integers,a < b。鉴于<代码>std:set<int> S,是有效的and elegant(最好没有明确的左轮)方式找到和储存(载于vector 页: 1 S。

Solution 1:

 vector<int> v;
 for(int i = a; i <= b; ++i)
 {
     if(S.find(i) == S.end())
     {
        v.push_back(i);
     }         
}

<<>Solution2:

缩略语

解决办法1 含有明确的循环,解决办法2 似乎效率不高(至少在记忆方面)。你会提出什么建议? 我正在寻找一种明智的STL-ish(boost也是可以接受的)辅助方式来做到这一点。

Answer 1

你可以做像你的解决办法2。但是,使用boost:irange, 它是数字范围的一个虚拟集装箱。这样,你就不会有明确的 lo,而是会步不前,不会留下太多的记忆。

使之更快,只包括使用<代码>的下限(<>限值/代码>/<代码>的下限/代码>/上限<<>代码/代码>确定的相关部分:

auto abRange = boost::irange(a,b+1);
std::set_difference(abRange.begin(), abRange.end(), 
                    s.lower_bound(a), s.upper_bound(b), 
                    std::back_inserter(resultVector));

Or using Boost.Range s set_difference:

boost::set_difference(boost::irange(a,b+1),
                      std::make_pair(s.lower_bound(a), s.upper_bound(b)),
                      std::back_inserter(resultVector));

Answer 2

下面的一段话可以避开,但我不敢肯定会给你留下什么:

void inSet(int i, int b, vector<int>& v, set<int>& S)
{
   if(S.find(i) == S.end())
        v.push_back(i);

   if(i<b)
        inSet(i+1,b,v,S);
}

// ... snip
vector<int> v;
inSet(a,b,v,S);

Also, is there not a loop putting all the integers [a,b] into a std::set in your solution 2?

Answer 3

http://www.un.org/Depts/DGACM/index_french.htm t 系指<代码>std:set——它简单是指一套逻辑;一组内容。如果对这两种收集进行分类,你可以简单地读到<条码>_intersection。这两辆集装箱到第三集装箱。

vector<int> common;
set_intersection(v.begin(), v.end(), s.begin(), s.end(), back_inserter(common));

EDIT:

下面是上述例子。这使用C++11 lambdas,但如果你没有C++11或能够使用斜体,你可以自行使用ctor子。指出缺乏明确的住所。

#include <set>
#include <vector>
#include <algorithm>
#include <iterator>
#include <functional>
#include <iostream>
using namespace std;

static const int numbers[] = {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181};
static const size_t num_numbers = sizeof(numbers)/sizeof(numbers[0]);

int main()
{
    /*** GET THE SET ****/
    set<int> s(begin(numbers), end(numbers));
    //copy(&numbers[0], &numbers[num_numbers], inserter(s, s.begin()));

    /*** GET THE NUMBERS TO LOOK FOR **/
    int first = 5, last = 10;
    vector<int> targets;
    generate_n(back_inserter(targets), last-first, [&first]() -> int {
        return first++;
    });

    /*** FIND THE INTERSECTION ***/
    vector<int> common;
    set_intersection(s.begin(), s.end(), targets.begin(), targets.end(), back_inserter(common));

    /*** DUMP RESULTS ***/
    cout << "Intersecton of:
	";
    copy(s.begin(), s.end(), ostream_iterator<int>(cout,"	"));
    cout << "
with:
	";
    copy(targets.begin(), targets.end(), ostream_iterator<int>(cout,"	"));
    cout << "
= = = = = = = =
	";
    copy(common.begin(), common.end(), ostream_iterator<int>(cout,"	"));
    cout << "
";

}

产出是:

Intersecton of:
        0       1       2       3       5       8       13      21      34
55      89      144     233     377     610     987     1597    2584    4181

with:
        5       6       7       8       9
= = = = = = = =
        5       8

Answer 4

您可以从<条码>S.down_有约束力(a)到<条码>,下限(b),并收集你没有发现的所有愤怒:

auto end = S.lower_bound(b);
int seen = a;

for (auto it = S.lower_bound(a); it < end; ++it) {
   for (int i = seen+1; i < *it; ++i)
      v.push_back(i);
   seen = *it;
}

It contains an explicit loop, but somehow you ll have to look at all the integers in [a,b].

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