题目地址：https://leetcode.com/problems/implement-queue-using-stacks/open in new window
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、 Youmustuseonlystandardoperationsofastack--whichmeansonlypushtotop,peek/popfromtop,size,andisemptyoperationsarevalid.；
2、 Dependingonyourlanguage,stackmaynotbesupportednatively.Youmaysimulateastackbyusingalistordeque(double-endedqueue),aslongasyouuseonlystandardoperationsofastack.；
3、 Youmayassumethatalloperationsarevalid(forexample,nopoporpeekoperationswillbecalledonanemptyqueue).；
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()
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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();
}
}
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AC:113ms
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