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).

題目意思：

　　用棧實現佇列

解題思路：

　　用兩個棧實現，一個棧用作佇列的出口，一個棧用作佇列的進口

擴充套件：

　　c++ 的STL中佇列的實現不是用的棧，其實現原理是開闢一段記憶體，該記憶體中每個節點指向一段連續的記憶體空間，然後維護這些結點，具體參考《STL原始碼剖析》，面試時可能會問到STL佇列的實現方式。STL中還有個優先權佇列，其實現原理是用的堆資料結構實現的，堆排序演算法最好自己能寫。

原始碼：

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();
    }
};