Implement Stack using Queues leetcode

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.

不用多想，可以使用两个queue来实现，代码如下

耗时在pop上

class Stack {
public:
    // Push element x onto stack.
    void push(int x) {
        que1.push(x);
    }    // Removes the element on top of the stack.
    void pop() {
        while (que1.size() > )
        {
            que2.push(que1.front());
            que1.pop();
        }
        que1.pop();
        while (!que2.empty())
        {
            que1.push(que2.front());
            que2.pop();
        }
    }    // Get the top element.
    int top() {
        return que1.back();
    }    // Return whether the stack is empty.
    bool empty() {
        return que1.empty();
    }
private:
    queue<int> que1;
    queue<int> que2;
};

查看其它人的代码，发现居然可以使用一个queue来实现，实现思路非常巧妙，每一次push操作都将queue之前的元素都移到后面，使其保持栈的排序

如：

push 1

1

push 2

2 1

push 3

3 2 1

class Stack {
public:
    queue<int> que;
    // Push element x onto stack.
    void push(int x) {
        que.push(x);
        for (int i = ; i<que.size() - ; ++i) {
            que.push(que.front());
            que.pop();
        }
    }    // Removes the element on top of the stack.
    void pop() {
        que.pop();
    }    // Get the top element.
    int top() {
        return que.front();
    }    // Return whether the stack is empty.
    bool empty() {
        return que.empty();
    }
};