716. Max Stack

• Difficulty: Easy

• Topics: Design

• Similar Questions:

  • Min 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:

Solutions:

class MaxStack {
public:
    /** initialize your data structure here. */
    MaxStack() {

    }

    void push(int x) {
        data.push(x);
        if (maxs.empty() || x >= maxs.top()) {
            maxs.push(x);
        }
    }

    int pop() {
        int val = data.top();
        data.pop();
        if (val == maxs.top()) {
            maxs.pop();
        }
        return val;
    }

    int top() {
        return data.top();
    }

    int peekMax() {
        return maxs.top();
    }

    int popMax() {
        int maxVal = maxs.top();
        stack<int> buffer;
        while (top() != maxVal) buffer.push(pop());
        pop();
        while (!buffer.empty()) {
            push(buffer.top());
            buffer.pop();
        }
        return maxVal;
    }
private:
    stack<int> data;
    stack<int> maxs;
};

/**
 * 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();
 */