题目描述

题目链接：225. Implement Stack using Queues

Implement a last in first out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal queue (push, top, pop, and empty).

Implement the MyStack class:

• void push(int x) Pushes element x to the top of the stack.
• int pop() Removes the element on the top of the stack and returns it.
• int top() Returns the element on the top of the stack.
• boolean empty() Returns true if the stack is empty, false otherwise.

例子

例子 1

Input: ["MyStack", "push", "push", "top", "pop", "empty"] [[], [1], [2], [], [], []] Output: [null, null, null, 2, 2, false] Explanation: MyStack myStack = new MyStack(); myStack.push(1); myStack.push(2); myStack.top(); // return 2 myStack.pop(); // return 2 myStack.empty(); // return False

Follow Up

Can you implement the stack such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer. You can use more than two queues.

Note

• 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, the queue may not be supported natively. You may simulate a queue using a list or deque (double-ended queue), as long as you use only a queue’s standard operations

Constraints

• 1 <= x <= 9
• At most 100 calls will be made to push, pop, top, and empty.
• All the calls to pop and top are valid.

解题思路

思路比较简单，由于栈是先入后出，因此每 push 一个元素我们要确保它在我们使用的 queue 的前面，因此只需要用一个临时 queue 将 x 压入，然后依次压入我们维护的 queue 的元素，最后将临时 queue 与我们维护的 queue 交换即可；其他操作可以直接使用 queue 的相应操作，代码如下：

#include <queue>

class MyStack {
public:
    /** Initialize your data structure here. */
    MyStack() {}

    /** Push element x onto stack. */
    void push(int x) {
        std::queue<int> temp;
        temp.push(x);
        while (!ordered_.empty()) {
            temp.push(ordered_.front());
            ordered_.pop();
        }
        ordered_.swap(temp);
    }

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

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

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

private:
    std::queue<int> ordered_;
};

• 时间复杂度:
  • push: O(n)
  • pop: O(1)
  • peek: O(1)
  • empty: O(1)
• 空间复杂度: O(n)

GitHub 代码同步地址： 225.ImplementStackUsingQueues.cpp

其他题目： GitHub: Leetcode-C++-Solution 博客： Leetcode-Solutions