Description
https://leetcode.com/problems/implement-queue-using-stacks/
Implement a last in first out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal queue (push
, top
, pop
, and empty
).
Implement the MyStack
class:
void push(int x)
Pushes element x to the top of the stack.int pop()
Removes the element on the top of the stack and returns it.int top()
Returns the element on the top of the stack.boolean empty()
Returnstrue
if the stack is empty,false
otherwise.
Notes:
- You must use only standard operations of a queue, which means only
push to back
,peek/pop from front
,size
, andis empty
operations are valid. - Depending on your language, the queue may not be supported natively. You may simulate a queue using a list or deque (double-ended queue), as long as you use only a queue’s standard operations.
Example 1:
Input ["MyStack", "push", "push", "top", "pop", "empty"] [[], [1], [2], [], [], []] Output
[null, null, null, 2, 2, false]
Explanation MyStack myStack = new MyStack(); myStack.push(1); myStack.push(2); myStack.top(); // return 2 myStack.pop(); // return 2 myStack.empty(); // return False
Constraints:
1 <= x <= 9
- At most
100
calls will be made topush
,pop
,top
, andempty
. - All the calls to
pop
andtop
are valid.
Follow-up: Can you implement the stack such that each operation is amortizedO(1)
time complexity? In other words, performing n
operations will take overall O(n)
time even if one of those operations may take longer. You can use more than two queues.
Explanation
Using two queues, one queue is for getting items, the other queue is only for internal conversion to get the last item.
Python Solution
import queue
class MyStack:
def __init__(self):
"""
Initialize your data structure here.
"""
self.queue1 = queue.Queue()
self.queue2 = queue.Queue()
def push(self, x: int) -> None:
"""
Push element x onto stack.
"""
self.queue1.put(x)
def pop(self) -> int:
"""
Removes the element on top of the stack and returns that element.
"""
while self.queue1.qsize() > 1:
self.queue2.put(self.queue1.get())
top = self.queue1.get()
self.queue1, self.queue2 = self.queue2, self.queue1
return top
def top(self) -> int:
"""
Get the top element.
"""
while self.queue1.qsize() > 1:
self.queue2.put(self.queue1.get())
top = self.queue1.get()
self.queue2.put(top)
self.queue1, self.queue2 = self.queue2, self.queue1
return top
def empty(self) -> bool:
"""
Returns whether the stack is empty.
"""
return not self.queue1.qsize()
# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()
- Time complexity: O(N).
- Space complexity: O(1).