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:
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 using only one queue?
import java.util.LinkedList;
import java.util.Queue;
public class MyStack {
private Queue<Integer> queueOne;
private Queue<Integer> queueTwo;
private int top;
// Initialize your data structure here.
public MyStack() {
queueOne = new LinkedList<>();
queueTwo = new LinkedList<>();
top = 0;
}
// Push element x onto stack.
public void push(int x) {
queueOne.add(x);
top = x;
}
// Removes the element on top of the stack and returns that element.
public int pop() {
while (queueOne.size() > 1) {
int val = queueOne.remove();
top = val;
queueTwo.add(val);
}
int popValue = queueOne.remove();
queueOne.addAll(queueTwo);
queueTwo.clear();
return popValue;
}
// Get the top element.
public int top() {
return top;
}
// Returns whether the stack is empty.
public boolean empty() {
return queueOne.isEmpty();
}
}