[英]Why is std::rotate faster than this way of doing it?
void rotate(vector <int> &a)
{
int lastElem = a[a.size()-1];
for(int i=a.size()-1;i>0;i--){
a[i] = a[i-1];
}
a[0] = lastElem;
}
Versus相对
rotate(a.begin(),a.end()-1,a.end());
As far as I can see the algorithm above is O(n) so why is STL way faster(I thought it was linear time as well).据我所见,上面的算法是 O(n) 那么为什么 STL 更快(我认为这也是线性时间)。
The standard library implementation of std::rotate
is likely using a call to memmove()
for bulk data copying. std::rotate
的标准库实现很可能使用调用memmove()
进行批量数据复制。 That would be one reason that it's faster than your hand-written loop.这将是它比您的手写循环更快的原因之一。
Since you are only rotating a single element you can replace your loop with a call tostd::copy_backward
.由于您只旋转单个元素,您可以通过调用
std::copy_backward
来替换循环。 That will also compile to memmove()
and provide better performance.这也将编译为
memmove()
并提供更好的性能。
void rotate(std::vector<int> &a)
{
int lastElem = a.back();
std::copy_backward(a.begin(), a.end() - 1, a.end()); // memmove()
a.front() = lastElem;
}
You can examine the generated assembly here on Compiler Explorer .您可以在 Compiler Explorer 上检查生成的程序集。
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