Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).
Implement the MyQueue class:
void push(int x) Pushes element x to the back of the queue.int pop() Removes the element from the front of the queue and returns it.int peek() Returns the element at the front of the queue.boolean empty() Returns true if the queue is empty, false otherwise.Notes:
push to top, peek/pop from top, size, and is empty operations are valid.
Example 1:
Input ["MyQueue", "push", "push", "peek", "pop", "empty"] [[], [1], [2], [], [], []] Output [null, null, null, 1, 1, false] Explanation MyQueue myQueue = new MyQueue(); myQueue.push(1); // queue is: [1] myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue) myQueue.peek(); // return 1 myQueue.pop(); // return 1, queue is [2] myQueue.empty(); // return false
Constraints:
1 <= x <= 9100 calls will be made to push, pop, peek, and empty.pop and peek are valid.
Follow-up: Can you implement the queue 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.
class MyQueue {
Stack<Integer> stk1 = new Stack<>();
Stack<Integer> stk2 = new Stack<>();
public MyQueue() {
}
public void push(int x) {
while (!stk1.isEmpty())
stk2.push(stk1.pop());
stk1.push(x);
}
public int pop() {
while (!stk2.isEmpty())
stk1.push(stk2.pop());
return stk1.pop();
}
public int peek() {
while (!stk2.isEmpty())
stk1.push(stk2.pop());
return stk1.peek();
}
public boolean empty() {
return stk1.isEmpty();
}
}
/**
* Your MyQueue object will be instantiated and called as such:
* MyQueue obj = new MyQueue();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.peek();
* boolean param_4 = obj.empty();
*/