# 232. Implement Queue using Stacks 用栈实现队列

@TOC

[LeetCode]

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).

## # 解题方法

### # Python解法

``````class MyQueue(object):

def __init__(self):
"""
"""
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

``````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