LeetCode 225. Implement Stack using Queues--用队列实现栈--C++解法

LeetCode 225. Implement Stack using Queues–用队列实现栈–C++解法

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LeetCode题解专栏：LeetCode题解
我做的所有的LeetCode的题目都放在这个专栏里，大部分题目Java和Python的解法都有。

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题目地址：Implement Stack using Queues - LeetCode

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Implement the following operations of a stack using queues.

push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
empty() -- Return whether the stack is empty.

Example:

MyStack stack = new MyStack();

stack.push(1);
stack.push(2);  
stack.top();   // returns 2
stack.pop();   // returns 2
stack.empty(); // returns false

Notes:

You must use only standard operations of a queue – which means only push to back, peek/pop from front, size, and is empty operations are valid.
Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

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这道题目是经典的用队列实现栈，虽然挺蠢的。
使用队列，入队或出队，至少有一个的时间复杂度要为O(n)。
使用链表的话可以做到O(1)。
C++解法如下：

class MyStack {
public:
    queue<int> que;

    /** Initialize your data structure here. */
    MyStack() {

    }

    /** Push element x onto stack. */
    void push(int x) {
        que.push(x);
        for (int i = 0; i < que.size() - 1; i++) {
            que.push(que.front());
            que.pop();
        }
    }

    /** Removes the element on top of the stack and returns that element. */
    int pop() {
        int front = top();
        que.pop();
        return front;
    }

    /** Get the top element. */
    int top() {
        return que.front();
    }

    /** Returns whether the stack is empty. */
    bool empty() {
        return que.empty();
    }
};

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack* obj = new MyStack();
 * obj->push(x);
 * int param_2 = obj->pop();
 * int param_3 = obj->top();
 * bool param_4 = obj->empty();
 */