# 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();
*/
``````