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Leetcode[232] Implement Queue using Stacks

29 Nov 2015

###Task1 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. 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).

###Python

class Queue(object):
    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack1 = []
        self.stack2 = []
        

    def push(self, x):
        """
        :type x: int
        :rtype: nothing
        """
        self.stack1.append(x)

    def pop(self):
        """
        :rtype: nothing
        """
        if len(self.stack2):
            return self.stack2.pop()
        else:
            while len(self.stack1):
                self.stack2.append(self.stack1.pop())
            return self.stack2.pop()
        

    def peek(self):
        """
        :rtype: int
        """
        if len(self.stack2):
            return self.stack2[-1]
        else:
            while len(self.stack1):
                self.stack2.append(self.stack1.pop())
            return self.stack2[-1]

    def empty(self):
        """
        :rtype: bool
        """
        return len(self.stack1) == 0 and len(self.stack2) == 0

####Single stack

class Queue:
    # initialize your data structure here.
    def __init__(self):
        self.stack = []

    # @param x, an integer
    # @return nothing
    def push(self, x):
        swap = []
        while self.stack:
            swap.append(self.stack.pop())
        swap.append(x)
        while swap:
            self.stack.append(swap.pop())

    # @return nothing
    def pop(self):
        self.stack.pop()

    # @return an integer
    def peek(self):
        return self.stack[-1]

    # @return an boolean
    def empty(self):
        return len(self.stack) == 0

###Java

class MyQueue {
    private Stack<Integer> stack1 = new Stack<Integer>();
    private Stack<Integer> stack2 = new Stack<Integer>();

    public void push(int x) {
        stack1.push(x);
    }
    
    public void pop() {
        if (stack2.isEmpty()) {
            while (!stack1.isEmpty()) {
                stack2.push(stack1.pop());   
            }
        }
        stack2.pop();
    }

    public int peek() {
        if (stack2.isEmpty()) {
            while (!stack1.isEmpty()) {
                stack2.push(stack1.pop());
            }
        }
        return stack2.peek();
    }

    public boolean empty() {
        return stack1.isEmpty() && stack2.isEmpty();
    }
}

###Points

Double Stack

双栈法：
维护两个栈inStack与outStack，其中inStack接收push操作新增的元素，outStack为pop/peek操作提供服务

由于栈具有后进先出（Last In First Out）的性质，栈A中的元素依次弹出并压入空栈B之后，栈A中元素的顺序会被逆转

当执行pop或者peek操作时，如果outStack中元素为空，则将inStack中的所有元素弹出并压入outStack，然后对outStack执行相应操作

由于元素至多只会从inStack向outStack移动一次，因此peek/pop操作的平摊开销为O(1)

Single stack

单栈法：
在执行push操作时，使用辅助栈swap，将栈中元素顺序按照push顺序的逆序存储。

此时，push操作的时间复杂度为O(n)，其余操作的时间复杂度为O(1)

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