716 Max Stack¶

Problem¶

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

Solutions:¶

```java class MaxStack { Stack data = new Stack(); Stack max = new Stack(); /** 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(); */ ```