LeetCode 225 Implement Stack using Queues（用队列来实现栈）（*）

翻译

[code]用队列来实现栈的如下操作。

push(x) —— 将元素x添加进栈
pop() —— 从栈顶移除元素
top() —— 返回栈顶元素
empty() —— 返回栈是否为空

注意：

你必须使用一个只有标准操作的队列。

也就是说，只有push/pop/size/empty等操作是有效的。

队列可能不被原生支持，这取决于你所用的语言。

只要你只是用queue的标准操作，你可以用list或者deque（double-ended queue）来模拟队列。

你可以假设所有的操作都是有效的（例如，pop或peek操作不会被用在空栈上）。

原文

[code]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.

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).

分析

代码

[code]class Stack {
public:
    queue<int> q, q2;

    // Push element x onto stack.
    void push(int x) {
        q.push(x);
    }

    // Removes the element on top of the stack.
    void pop() {
        if (q.size() == 1) q.pop();
        else {
            while (q.size() > 1) {
                q2.push(q.front());
                q.pop();
            }
            q.pop();
            while (q2.size() > 0) {
                q.push(q2.front());
                q2.pop();
            }
        }
    }

    // Get the top element.
    int top() {
        if (q.size() == 1) return q.front();
        while (q.size() > 1) {
            q2.push(q.front());
            q.pop();
        }
        int temp = q.front();
        q2.push(q.front());
        q.pop();
        while (q2.size() > 0) {
            q.push(q2.front());
            q2.pop();
        }
        return temp;
    }

    // Return whether the stack is empty.
    bool empty() {
        return q.empty();
    }
};