LeetCode 232： 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 onlypush 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).

题目要求通过栈来模拟队列的行为。与此类似的还有通过队列模拟栈（http://blog.csdn.net/sunao2002002/article/details/46482673），此题是算法导论第十章的一道题。算法如下：

堆栈a和b，a用作入队，b出队

（1）判队满：如果a满且b不为空，则队满

（2）判队空：如果a和b都为空，则队空

（3）入队：首先判队满。

    若队不满：（1）栈a若不满，则直接压入栈a

                        (2)若a满，则将a中的所有元素弹出到栈b中，然后再将元素入栈a

（4）出队：(1)若b空就将a中的所有元素弹出到栈b中，然后出栈

                      （2）b不空就直接从b中弹出元素

代码如下：

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