Implement a last in first out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal queue (
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.
trueif the stack is empty,
- You must use only standard operations of a queue, which means only
push to back,
peek/pop from front,
is emptyoperations 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.
Input ["MyStack", "push", "push", "top", "pop", "empty"] [, , , , , ] 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
1 <= x <= 9
- At most
100calls will be made to
- All the calls to
Follow-up: Can you implement the stack such that each operation is amortized
O(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.
Solution in python
class MyStack: def __init__(self): """ Initialize your data structure here. """ self.stack = deque() def push(self, x: int) -> None: """ Push element x onto stack. """ self.stack.append(x) def pop(self) -> int: """ Removes the element on top of the stack and returns that element. """ return self.stack.pop() def top(self) -> int: """ Get the top element. """ return self.stack[-1] def empty(self) -> bool: """ Returns whether the stack is empty. """ return False if self.stack else True # 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()