Leetcode61: Implement Queue using Stacks

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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 toppeek/pop from topsize, 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。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

class Queue {public:    stack<int> in ;    stack<int> out;    // Push element x to the back of queue.    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();    }};


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