[英]return top 5 value using vector iterator
大家好,有什么方法可以使用向量迭代器返回向量中的前5个元素吗?
在此示例中,我将仅获得向量本身中的所有值。
// vector::begin/end
#include <iostream>
#include <vector>
int main ()
{
std::vector<int> myvector;
for (int i=1; i<=10; i++) myvector.push_back(i);
std::cout << "myvector contains:";
for (std::vector<int>::iterator it = myvector.begin() ; it != myvector.end(); ++it)
std::cout << ' ' << *it;
std::cout << '\n';
return 0;
}
嗯,感谢您的所有迅速答复。但是,当我尝试将它们放入函数中时,为什么会出现编译错误?
void Test::topfives()
{
topfive.assign( point1.begin(), point1.end() );
sort(topfive.begin(), topfive.end(), sortByCiv);
}
void Test::DisplayTopFiveResult()
{
test.topfives();
copy(topfive.begin(), topfive.begin()+ min(topfive.size(), (size_t )5),
ostream_iterator<Level>(cout << level.displayClassresult()));
}
将myvector.begin()
前进5
std::copy(myvector.begin(),
myvector.begin()+std::min(myvector.size(), (size_t)5),
std::ostream_iterator<int>(std::cout,"\n"));
这从myvector
最多打印前5个元素
看这里
#include <iostream>
#include <vector>
int main ()
{
std::vector<int> myvector;
for (int i=1; i<=10; i++) myvector.push_back(i);
std::cout << "myvector contains:";
std::vector<int>::iterator it = myvector.begin()
for (int i = 0; i < 5 && it != myvector.end(); i++) {
std::cout << ' ' << *it;
++it;
}
std::cout << '\n';
return 0;
}
std :: vector支持随机访问迭代器 ,这意味着您可以执行以下操作:
myvector.begin() + 5
因此,您的代码有两个选择:
std::cout << "myvector contains:";
size_t maxElements = std::min(myvector.size(), size_t(5));
for (size_t i = 0; i < numElements; ++i) {
std::cout << ' ' << myvector[i];
要么
auto startIt = myvector.begin();
auto endIt = myvector.begin() + 5;
for (auto it = startIt; it != endIt; ++it)
std::cout << ' ' << *it;
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