232. Implement Queue using Stacks 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.

Example:

MyQueue queue = new MyQueue();

queue.push(1);
queue.push(2);  
queue.peek();  // returns 1
queue.pop();   // returns 1
queue.empty(); // returns false

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). 232. Implement Queue using Stacks

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本题要求用堆栈来实现队列的功能。因为堆栈是后进先出，而队列是先进先出，顺序正好相反。比如进入队列的顺序是1,2,3，但是进到堆栈stk1之后，从栈顶到栈底的顺序就变成了3,2,1，此时如果要实现队列出队的效果，就必须借助另一个堆栈stk2，把stk1的内容压入到stk2，这样从栈顶到栈底的顺序就和队列一样了，是1,2,3。 所以本题借助两个堆栈stk1和stk2，stk1就是常规的数据来了就压栈，当需要实现队列的pop效果时，就借助stk2倒腾一下，如果又要实现队列的push效果时，又要把stk2的数据捣腾回stk1。 总的来说，我们使用懒人规则，每次直到调用某个操作，且当前stk*不能满足要求时，才捣腾stk1和stk2。 完整代码如下：

class MyQueue {
private:
    stack<int> stk1, stk2; // 输入来源1,2,3，stk1从栈底到栈顶的顺序是1,2,3，stk2从栈底到栈顶的顺序是3,2,1
public: /** Initialize your data structure here. */
    MyQueue() {} /** Push element x to the back of queue. */
    void push(int x)
    {
        while (!stk2.empty()) {
            stk1.push(stk2.top());
            stk2.pop();
        }
        stk1.push(x);
    } /** Removes the element from in front of queue and returns that element. */
    int pop()
    {
        while (!stk1.empty()) {
            stk2.push(stk1.top());
            stk1.pop();
        }
        int top = stk2.top();
        stk2.pop();
        return top;
    } /** Get the front element. */
    int peek()
    {
        while (!stk1.empty()) {
            stk2.push(stk1.top());
            stk1.pop();
        }
        return stk2.top();
    } /** Returns whether the queue is empty. */
    bool empty() { return stk1.empty() && stk2.empty(); }
};

本代码提交AC，用时3MS。
二刷。其实push的时候可以不需要把stk2的数据倒腾回stk1，stk2的数据始终是队列头的，stk1的数据始终是队列尾的，代码如下：

class MyQueue {
private:
    stack<int> stk1, stk2;
    void adjust()
    {
        while (!stk1.empty()) {
            stk2.push(stk1.top());
            stk1.pop();
        }
    }
public: /** Initialize your data structure here. */
    MyQueue() {} /** Push element x to the back of queue. */
    void push(int x) { stk1.push(x); } /** Removes the element from in front of queue and returns that element. */
    int pop()
    {
        if (stk2.empty())
            adjust();
        int ans = stk2.top();
        stk2.pop();
        return ans;
    } /** Get the front element. */
    int peek()
    {
        if (stk2.empty())
            adjust();
        int ans = stk2.top();
        return ans;
    } /** Returns whether the queue is empty. */
    bool empty() { return stk1.empty() && stk2.empty(); }
};

本代码提交AC，用时3MS。