【数据结构入门

原题链接：Leetcode 232. Implement Queue using Stacks

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

• void push(int x) Pushes element x to the back of the queue.
• int pop() Removes the element from the front of the queue and returns it.
• int peek() Returns the element at the front of the queue.
• boolean empty() Returns true if the queue is empty, false otherwise.

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

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false

Constraints:

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

Follow-up:

Can you implement the queue 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.

栈是先入后出的，队列是先入先出的。
用两个栈（输入栈和输出栈）通过倒转输入的序列来实现队列

class MyQueue {
private:
    // 一个输入栈 一个输出栈
    stack inStack, outStack;

    // 将输入栈的元素依次添加到输出栈中
    void transfer() {
        while (!inStack.empty()) {
            outStack.push(inStack.top());
            inStack.pop();
        }
    }

public:
    MyQueue() {}

    // 入队
    void push(int x) {
        inStack.push(x);
    }

    // 出队
    int pop() {
        if (outStack.empty()) {
            transfer();
        }
        int x = outStack.top();
        outStack.pop();
        return x;
    }

    // 返回队头元素
    int peek() {
        if (outStack.empty()) {
            transfer();
        }
        return outStack.top();
    }

    // 判断是否为空
    bool empty() {
        return inStack.empty() && outStack.empty();
    }
};

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