Question

Design a max stack that supports push, pop, top, peekMax and popMax.

push(x) – Push element x onto stack. pop() – Remove the element on top of the stack and return it. top() – Get the element on the top. peekMax() – Retrieve the maximum element in the stack. popMax() – Retrieve the maximum element in the stack, and remove it. If you find more than one maximum elements, only remove the top-most one.

Example 1:
MaxStack stack = new MaxStack();
stack.push(5);
stack.push(1);
stack.push(5);
stack.top(); -> 5
stack.popMax(); -> 5
stack.top(); -> 1
stack.peekMax(); -> 5
stack.pop(); -> 1
stack.top(); -> 5

Note:

  1. -1e7 <= x <= 1e7
  2. Number of operations won’t exceed 10000.
  3. The last four operations won’t be called when stack is empty.

Solutions

  • Method 1: LinkedList + PriorityQueue 
    class MaxStack {
    private PriorityQueue<Integer> pq;
    private LinkedList<Integer> values;
    /** initialize your data structure here. */
    public MaxStack() {
        this.pq = new PriorityQueue<>(new Comparator<Integer>(){
            @Override
            public int compare(Integer a, Integer b){
                return b - a;
            }
        });
        this.values = new LinkedList<>();
    }

    public void push(int x) {
        this.values.addFirst(x);
        this.pq.offer(x);
    }

    public int pop() {
        Integer first = this.values.pollFirst();
        pq.remove(first);
        return first;
    }

    public int top() {
        return this.values.get(0);
    }

    public int peekMax() {
        return this.pq.peek();
    }

    public int popMax() {
        Integer max = pq.poll();
        this.values.remove(max);
        return max;
    }
}

/**
 * Your MaxStack object will be instantiated and called as such:
 * MaxStack obj = new MaxStack();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.top();
 * int param_4 = obj.peekMax();
 * int param_5 = obj.popMax();
 */