A classic interview question. This link 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<int> stack1; 38 stack<int> stack2; 39 };