716 Max Stack

Problem

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

1. push(x) -- Push element x onto stack.
2. pop() -- Remove the element on top of the stack and return it.
3. top() -- Get the element on the top.
4. peekMax() -- Retrieve the maximum element in the stack.
5. 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: -1e7 <= x <= 1e7 Number of operations won't exceed 10000. The last four operations won't be called when stack is empty.

Solutions:

class MaxStack {
    Stack<Integer> data = new Stack<Integer>();
    Stack<Integer> max = new Stack<Integer>();
    /** initialize your data structure here. */
    public MaxStack() {

    }
    //O(1);
    public void push(int x) {
        data.push(x);
        if (max.isEmpty()) {
            max.push(x);
        }
        else {
            max.push(Math.max(x, max.peek()));
        }
    }
    //O(1);
    public int pop() {
        max.pop();
        return data.pop();
    }
    //O(1);
    public int top() {
        return data.peek();
    }
    //O(1);
    public int peekMax() {
        return max.peek();
    }
    //O(n);
    public int popMax() {
        int res = max.peek();
        Stack<Integer> tmp = new Stack<Integer>();
        while (data.peek() != res) {
            tmp.push(data.pop());
            max.pop();
        }
        data.pop();
        max.pop();
        while (!tmp.isEmpty()) {
            push(tmp.pop());
        }
        return res;
    }
}

/**
 * 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();
 */