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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()
Returns true
if the stack is empty, false
otherwise.Notes:
push to back
, peek/pop from front
, size
, and is empty
operations are valid.
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
100
calls will be made to push
, pop
, top
, and empty
.pop
and top
are valid.
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.
class MyStack:
def __init__(self):
"""
Initialize your data structure here.
"""
self.q = []
def push(self, x: int) -> None:
"""
Push element x onto stack.
"""
self.q.append(x)
n = len(self.q)
for i in range(1, n):
self.q.append(self.q.pop(0))
def pop(self) -> int:
"""
Removes the element on top of the stack and returns that element.
"""
return self.q.pop(0)
def top(self) -> int:
"""
Get the top element.
"""
return self.q[0]
def empty(self) -> bool:
"""
Returns whether the stack is empty.
"""
return len(self.q) == 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()
class MyStack {
private Deque<Integer> q;
/** Initialize your data structure here. */
public MyStack() {
q = new ArrayDeque<>();
}
/** Push element x onto stack. */
public void push(int x) {
q.offerLast(x);
int n = q.size();
while (n-- > 1) {
q.offerLast(q.pollFirst());
}
}
/** Removes the element on top of the stack and returns that element. */
public int pop() {
return q.pollFirst();
}
/** Get the top element. */
public int top() {
return q.peekFirst();
}
/** Returns whether the stack is empty. */
public boolean empty() {
return q.isEmpty();
}
}
/**
* Your MyStack object will be instantiated and called as such:
* MyStack obj = new MyStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* boolean param_4 = obj.empty();
*/
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