225. Implement Stack using Queues

Easy


Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (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() Returns true if the stack is empty, false otherwise.

Notes:

  • You must use only standard operations of a queue, which means that only push to back, peek/pop from front, size and is 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 to push, pop, top, and empty.
  • All the calls to pop and top are valid.

 

Follow-up: Can you implement the stack using only one queue?





 class MyStack:
    """
    Using One Queue
    """

    def __init__(self):
        self.q1 = []
        self.top_val = None

    def push(self, x: int) -> None:
        self.q1.append(x)
        self.top_val = x


    def pop(self) -> int:
        size  = len(self.q1)
        while size > 1:
            item = self.q1.pop(0)
            self.q1.append(item)
            self.top_val = item
            size = size - 1

        return self.q1.pop(0)

    def top(self) -> int:
        return self.top_val

    def empty(self) -> bool:
        return len(self.q1)==0


#     """
#     Using Two Queue
#     """

#     def __init__(self):
#         self.q1 = []
#         self.q2 = []
#         self.top_val = None

#     def push(self, x: int) -> None:
#         self.q1.append(x)
#         self.top_val = x


#     def pop(self) -> int:

#         while len(self.q1) > 1:
#             # pop element and add to q2
#             item = self.q1.pop(0)
#             self.top_val = item
#             self.q2.append(item)
#         # performing stack pop
#         val = self.q1.pop()
#         # copying q2 to q1
#         self.q1 = self.q2.copy()
#         self.q2 = []
#         return val

#     def top(self) -> int:
#         return self.top_val

#     def empty(self) -> bool:
#         return len(self.q1) == 0


# 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()

Random Note


leetcode problem 287. Find the Duplicate Number has 7 solving approach. WOW!!!