LeetCode题解232.implement-queue-using-stacks

题目地址

https://leetcode.com/problems/implement-queue-using-stacks/description/

题目描述

Implement the following operations of a queue using stacks.

push(x) -- Push element x to the back of queue.
pop() -- Removes the element from in front of queue.
peek() -- Get the front element.
empty() -- Return whether the queue is empty.
Example:

MyQueue queue = new MyQueue();

queue.push(1);
queue.push(2);
queue.peek();  // returns 1
queue.pop();   // returns 1
queue.empty(); // returns false
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, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

思路

这道题目是让我们用栈来模拟实现队列。我们知道栈和队列都是一种受限的数据结构。
栈的特点是只能在一端进行所有操作，队列的特点是只能在一端入队，另一端出队。

在这里我们可以借助另外一个栈，也就是说用两个栈来实现队列的效果。这种做法的时间复杂度和空间复杂度都是O(n)。

由于栈只能操作一端，因此我们peek或者pop的时候也只去操作顶部元素，要达到目的
我们需要在push的时候将队头的元素放到栈顶即可。

因此我们只需要在push的时候，用一下辅助栈即可。
具体做法是先将栈清空并依次放到另一个辅助栈中，辅助栈中的元素再次放回栈中，最后将新的元素push进去即可。

比如我们现在栈中已经是1，2，3，4了。我们现在要push一个5.

push之前是这样的：

LeetCode题解232.implement-queue-using-stacks

然后我们将栈中的元素转移到辅助栈：

LeetCode题解232.implement-queue-using-stacks

最后将新的元素添加到栈顶。

LeetCode题解232.implement-queue-using-stacks

整个过程是这样的：

LeetCode题解232.implement-queue-using-stacks

关键点解析

在push的时候利用辅助栈(双栈)

代码

语言支持：JS, Python, Java

Javascript Code:

/*
 * @lc app=leetcode id=232 lang=javascript
 *
 * [232] Implement Queue using Stacks
 */
/**
 * Initialize your data structure here.
 */
var MyQueue = function() {
  // tag: queue stack array
  this.stack = [];
  this.helperStack = [];
};

/**
 * Push element x to the back of queue.
 * @param {number} x
 * @return {void}
 */
MyQueue.prototype.push = function(x) {
  let cur = null;
  while ((cur = this.stack.pop())) {
    this.helperStack.push(cur);
  }
  this.helperStack.push(x);

  while ((cur = this.helperStack.pop())) {
    this.stack.push(cur);
  }
};

/**
 * Removes the element from in front of queue and returns that element.
 * @return {number}
 */
MyQueue.prototype.pop = function() {
  return this.stack.pop();
};

/**
 * Get the front element.
 * @return {number}
 */
MyQueue.prototype.peek = function() {
  return this.stack[this.stack.length - 1];
};

/**
 * Returns whether the queue is empty.
 * @return {boolean}
 */
MyQueue.prototype.empty = function() {
  return this.stack.length === 0;
};

/**
 * Your MyQueue object will be instantiated and called as such:
 * var obj = new MyQueue()
 * obj.push(x)
 * var param_2 = obj.pop()
 * var param_3 = obj.peek()
 * var param_4 = obj.empty()
 */

Python Code:

class MyQueue:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.stack = []
        self.help_stack = []

    def push(self, x: int) -> None:
        """
        Push element x to the back of queue.
        """
        while self.stack:
            self.help_stack.append(self.stack.pop())
        self.help_stack.append(x)
        while self.help_stack:
            self.stack.append(self.help_stack.pop())

    def pop(self) -> int:
        """
        Removes the element from in front of queue and returns that element.
        """
        return self.stack.pop()

    def peek(self) -> int:
        """
        Get the front element.
        """
        return self.stack[-1]

    def empty(self) -> bool:
        """
        Returns whether the queue is empty.
        """
        return not bool(self.stack)


# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()

Java Code

class MyQueue {
    Stack<Integer> pushStack = new Stack<> ();
    Stack<Integer> popStack = new Stack<> ();

    /** Initialize your data structure here. */
    public MyQueue() {

    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        while (!popStack.isEmpty()) {
            pushStack.push(popStack.pop());
        }
        pushStack.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        while (!pushStack.isEmpty()) {
            popStack.push(pushStack.pop());
        }
        return popStack.pop();
    }
    
    /** Get the front element. */
    public int peek() {
        while (!pushStack.isEmpty()) {
            popStack.push(pushStack.pop());
        }
        return popStack.peek();
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        return pushStack.isEmpty() && popStack.isEmpty();
    }
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue obj = new MyQueue();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.peek();
 * boolean param_4 = obj.empty();
 */

扩展

类似的题目有用队列实现栈，思路是完全一样的，大家有兴趣可以试一下。
栈混洗也是借助另外一个栈来完成的，从这点来看，两者有相似之处。

LeetCode题解232.implement-queue-using-stacks

题目地址

题目描述

思路

关键点解析

代码

扩展

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