A classic interview question. has a nice explanation of the idea using two stacks and its amortized time complexity.

I use the same idea in my code, which is as follows.

1 class Queue { 2 public: 3     // Push element x to the back of queue. 4     void push(int x) { 5         stack1.push(x); 6     } 7  8     // Removes the element from in front of queue. 9     void pop(void) {10         if (stack2.empty()) {11             while (!stack1.empty()) {12                 int elem = stack1.top();13                 stack1.pop();14                 stack2.push(elem);15             }16         }17         stack2.pop();18     }19 20     // Get the front element.21     int peek(void) {22         if (stack2.empty()) {23             while (!stack1.empty()) {24                 int elem = stack1.top();25                 stack1.pop();26                 stack2.push(elem);27             }28         }29         return stack2.top();30     }31 32     // Return whether the queue is empty.33     bool empty(void) {34         return stack1.empty() && stack2.empty();35     }36 private:37     stack
     
       stack1;38     stack
      
        stack2;39 };