Leetcode Implement Queue using Stacks

Implement the following operations of a queue using stacks.

• push(x) -- Push element x to the back of queue.
• pop() -- Removes the element from in front of queue.
• peek() -- Get the front element.
• empty() -- Return whether the queue is empty.

Notes:

• You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is empty operations are valid.
• Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
• You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

题目意思：

　　用栈实现队列

解题思路：

　　用两个栈实现，一个栈用作队列的出口，一个栈用作队列的进口

扩展：

　　c++ 的STL中队列的实现不是用的栈，其实现原理是开辟一段内存，该内存中每个节点指向一段连续的内存空间，然后维护这些结点，具体参考《STL源码剖析》，面试时可能会问到STL队列的实现方式。STL中还有个优先权队列，其实现原理是用的堆数据结构实现的，堆排序算法最好自己能写。

源代码：

 class Queue{
     stack<int> inStack, outStack;
 public:
     void  push(int x){
         inStack.push(x);
     }
     void pop(void){
         if(outStack.empty()){
             while(!inStack.empty()){
                 outStack.push(inStack.top());
                 inStack.pop();
             }
         }
         outStack.pop();
     }

     int peek(void){
         if(outStack.empty()){
             while(!inStack.empty()){
                 outStack.push(inStack.top());
                 inStack.pop();
             }
         }
         return outStack.top();
     }

     bool empty(void) {
         return inStack.empty() && outStack.empty();
     }
 };