Description:

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.

Link:

https://leetcode.com/problems/implement-queue-using-stacks/#/description

解題方法：

用兩個棧in和out來解決，當(dāng)調(diào)用函數(shù)pop()和peek()時，當(dāng)out為空時，把in棧的元素壓到out里面去，然后對out進(jìn)行pop()和top()的調(diào)用。

Time Complexity：

pop()和peek()：O(1)到 O(N)

完整代碼：

class MyQueue 
{
private:
    stack<int> inS;
    stack<int> outS;
    void inToOut()
    {
        while(!inS.empty())
        {
            outS.push(inS.top());
            inS.pop();
        }
    }
public:
    /** Push element x to the back of queue. */
    void push(int x) 
    {
        inS.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() 
    {
        if(outS.empty())
            inToOut();
        int front = outS.top();
        outS.pop();
        return front;
    }
    
    /** Get the front element. */
    int peek() 
    {
        if(outS.empty())
            inToOut();
        return outS.top();
    }
    
    /** Returns whether the queue is empty. */
    bool empty() 
    {
        return inS.empty() && outS.empty();
    }
};