232, 用栈实现队列

一、题目描述

请你仅使用两个栈实现先入先出队列。队列应当支持一般队列支持的所有操作（push、pop、peek、empty）。

实现MyQueue类：

• void push(int x)将元素x推到队列的末尾
• int pop()从队列的开头移除并返回元素
• int peek()返回队列开头的元素
• boolean empty()如果队列为空，返回true；否则，返回false

说明:

• 你只能使用标准的栈操作——也就是只有push to top, peek/pop from top, size, 和is empty操作是合法的。
• 你所使用的语言也许不支持栈。你可以使用list或者deque（双端队列）来模拟一个栈，只要是标准的栈操作即可。

示例 1
输入:
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
输出:
[null, null, null, 1, 1, false]
解释:

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

提示

• 1 <= x <= 9
• 最多调用100次push、pop、peek和empty
• 假设所有操作都是有效的（例如，一个空的队列不会调用pop或者peek操作）

进阶
你能否实现每个操作均摊时间复杂度为O(1)的队列？换句话说，执行n个操作的总时间复杂度为O(n)，即使其中一个操作可能花费较长时间。

相关主题

• 栈
• 设计
• 队列

二、题解

pub struct MyQueue {
    s1: Vec<i32>,
    s2: Vec<i32>,
}

impl MyQueue {
    pub fn new() -> Self {
        MyQueue {
            s1: vec![],
            s2: vec![],
        }
    }
}

public class MyQueue {
    private Deque<Integer> s1;
    private Deque<Integer> s2;

    public MyQueue() {
        this.s1 = new ArrayDeque<>();
        this.s2 = new ArrayDeque<>();
    }
}

方法 1: 双栈法

impl MyQueue {
    /// Time Complexity: O(n)
    ///
    /// Space Complexity: O(n)
    pub fn push(&mut self, x: i32) {
        while let Some(val) = self.s1.pop() {
            self.s2.push(val);
        }
        self.s1.push(x);
        while let Some(val) = self.s2.pop() {
            self.s1.push(val);
        }
    }

    /// Time Complexity: O(1)
    ///
    /// Space Complexity: O(1)
    pub fn pop(&mut self) -> i32 {
        self.s1.pop().unwrap()
    }

    /// Time Complexity: O(1)
    ///
    /// Space Complexity: O(1)
    pub fn peek(&mut self) -> i32 {
        *self.s1.last().unwrap()
    }

    /// Time Complexity: O(1)
    ///
    /// Space Complexity: O(1)
    pub fn empty(&self) -> bool {
        self.s1.is_empty()
    }
}

class MyQueue {
    // Time Complexity: O(n)
    //
    // Space Complexity: O(n)
    public void push(int x) {
        while (!this.s1.isEmpty()) {
            this.s2.push(this.s1.pop());
        }
        this.s1.push(x);
        while (!this.s2.isEmpty()) {
            this.s1.push(this.s2.pop());
        }
    }

    // Time Complexity: O(1)
    //
    // Space Complexity: O(1)
    public int pop() {
        return this.s1.pop();
    }

    // Time Complexity: O(1)
    //
    // Space Complexity: O(1)
    public int peek() {
        return this.s1.peek();
    }

    // Time Complexity: O(1)
    //
    // Space Complexity: O(1)
    public boolean empty() {
        return this.s1.isEmpty();
    }
}

方法 2: 优化双栈法

impl MyQueue {
    /// Time Complexity: O(1)
    ///
    /// Space Complexity: O(n)
    pub fn push(&mut self, x: i32) {
        self.s1.push(x);
    }

    /// Time Complexity: Amortized O(1), Worst-case O(n)
    ///
    /// Space Complexity: O(1)
    pub fn pop(&mut self) -> i32 {
        if self.s2.is_empty() {
            while let Some(val) = self.s1.pop() {
                self.s2.push(val);
            }
        }
        self.s2.pop().unwrap()
    }

    /// Time Complexity: O(1)
    ///
    /// Space Complexity: O(1)
    pub fn peek(&mut self) -> i32 {
        if self.s2.is_empty() {
            *self.s1.first().unwrap()
        } else {
            *self.s2.last().unwrap()
        }
    }

    /// Time Complexity: O(1)
    ///
    /// Space Complexity: O(1)
    pub fn empty(&self) -> bool {
        self.s1.is_empty() && self.s2.is_empty()
    }
}

class MyQueue {
    // Time Complexity: O(1)
    //
    // Space Complexity: O(1)
    public void push(int x) {
        this.s1.push(x);
    }

    // Time Complexity: Amortized O(1), Worst-case O(n)
    //
    // Space Complexity: O(1)
    public int pop() {
        if (this.s2.isEmpty()) {
            while (!this.s1.isEmpty()) {
                this.s2.push(this.s1.pop());
            }
        }
        return this.s2.pop();
    }

    // Time Complexity: O(1)
    //
    // Space Complexity: O(1)
    public int peek() {
        if (this.s2.isEmpty()) {
            return this.s1.peekLast();
        } else {
            return this.s2.peek();
        }
    }

    // Time Complexity: O(1)
    //
    // Space Complexity: O(1)
    public boolean empty() {
        return this.s1.isEmpty() && this.s2.isEmpty();
    }
}