【LeetCode】232. Implement Queue using Stacks 解题报告（Python & Java）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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目录

• • 题目描述
  • 题目大意
  • 解题方法
  • • Python解法
    • Java解法
  • 日期

[LeetCode]

题目地址：https://leetcode.com/problems/implement-queue-using-stacks/

Total Accepted: 42648 Total Submissions: 125482 Difficulty: Easy

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:

1. 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.
2. 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.
3. You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

题目大意

使用栈来实现一个队列。

解题方法

众所周知，需要用两个栈。只要想清楚两个栈来左右翻转就好了。

Python解法

下面的python代码是把stack2当做是和队列顺序一样的，这样的话，如果stack2不空，那么久弹出元素就行。否则，如果stack1中有元素，那么在做push和pop的时候，需要先把stack1中的元素颠倒到stack2中。

class MyQueue(object):

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.stack1 = []
        self.stack2 = []

    def push(self, x):
        """
        Push element x to the back of queue.
        :type x: int
        :rtype: void
        """

        self.stack1.append(x)

    def pop(self):
        """
        Removes the element from in front of queue and returns that element.
        :rtype: int
        """
        if self.stack2:
            return self.stack2.pop()
        else:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
            return self.stack2.pop()

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

    def empty(self):
        """
        Returns whether the queue is empty.
        :rtype: bool
        """
        return not self.stack1 and not self.stack2

# 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解法

注意，A栈的元素顺序和队列的元素顺序是一样的。也就是说当pop()或者peek()的时候，其实直接把最上面的元素给拿出来就好了。

class MyQueue {
	Stack<Integer> stackA = new Stack<Integer>();
	Stack<Integer> stackB = new Stack<Integer>();

	// Push element x to the back of queue.
	public void push(int x) {
		if (stackA.isEmpty()) {
			stackA.push(x);
			System.out.println(stackA.toString());
			return;
		}
		while (!stackA.isEmpty()) {
			stackB.push(stackA.pop());
		}
		stackB.push(x);
		while (!stackB.isEmpty()) {
			stackA.push(stackB.pop());
		}
		System.out.println(stackA.toString());
	}

	// Removes the element from in front of queue.
	public void pop() {
		stackA.pop();
		System.out.println(stackA.toString());
	}

	// Get the front element.
	public int peek() {
		return stackA.peek();
	}

	// Return whether the queue is empty.
	public boolean empty() {
		return stackA.isEmpty();
	}
}

AC:113ms

日期

2016 年 05月 8日
2018 年 11 月 21 日 —— 又是一个美好的开始