LeetCode :232. Implement Queue using Stacks(剑指offer面试题)

问题描述：

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).

分析：

队列：先进先出

堆栈：后进先出

定义两个栈stack1、stack2，新元素进队列时，任选一个堆栈（这里选stack1），出队列时，要出最先进入队列的元素，由于队列式先进先出，所以要把stack1中已有的元素依次出栈，在进栈放入stack2,。

注意点：新元素进队列选取的stack1，则当stack2不为空是，stack2中的栈定元素必定是最先进入队列的元素，因此不论是出队列还是求队首元素，首先要判断stack2是否为空！！

AC代码：

class Queue {public:    // Push element x to the back of queue.    stack<int>stack1;    stack<int>stack2;    void push(int x)     {        stack1.push(x);    }    // Removes the element from in front of queue.    void pop(void)    {        if(!stack2.empty())            stack2.pop();        else        {            while(!stack1.empty())            {                int temp = stack1.top();                stack1.pop();                stack2.push(temp);            }            stack2.pop();        }            }    // Get the front element.    int peek(void)    {        if(!stack2.empty())            return stack2.top();        else        {            while(!stack1.empty())            {                int temp = stack1.top();                stack1.pop();                stack2.push(temp);            }              return stack2.top();        }            }    // Return whether the queue is empty.    bool empty(void)     {        if(stack1.empty() && stack2.empty())            return true;        else            return false;    }};