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.
- You must use only standard operations of a stack -- which means only
push to top
,peek/pop from top
,size
, andis 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。算法例如以下:
堆栈a和b。a用作入队,b出队
(1)判队满:假设a满且b不为空,则队满
(2)判队空:假设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中的全部元素弹出到栈b中,然后再将元素入栈a
(4)出队:(1)若b空就将a中的全部元素弹出到栈b中。然后出栈
(2)b不空就直接从b中弹出元素
代码例如以下: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(); } };