倒排队列的渐近分析(Asymptotic Analysis of Reversing a Queue)

    void reverseQueue(queue<int>& Queue)
    {
         stack<int> Stack;
         while (!Queue.empty())
         {
             Stack.push(Queue.front());
             Queue.pop();
         }
         while (!Stack.empty())
         {
             Queue.push(Stack.top());
             Stack.pop();
         }
    }
 
 我想知道这个函数的Big-O或Big-Theta符号是什么，如果我们用n个元素的队列来调用它的话。 它会是沿着O（n ^ 2）行的东西，因为我们推动并弹出n个元素两次，以便将它从栈中移回到队列中，并且以相反的顺序排列？ 感谢您的任何帮助。 

    void reverseQueue(queue<int>& Queue)
    {
         stack<int> Stack;
         while (!Queue.empty())
         {
             Stack.push(Queue.front());
             Queue.pop();
         }
         while (!Stack.empty())
         {
             Queue.push(Stack.top());
             Stack.pop();
         }
    }
 
I was wondering what the Big-O or Big-Theta notation of this function would be, if we called it with a Queue of n elements. Would it be something along the lines of O(n^2), since we're pushing and popping n elements twice in order to move it from the stack back to the queue in a reversed order? Thank you for any help.

原文：https://stackoverflow.com/questions/50283059