Problem

Implement a stack with min() function, which will return the smallest number in the stack.

It should support push, pop and min operation all in O(1) cost.

Example

push(1)
pop() // return 1
push(2)
push(3)
min() // return 2
push(1)
min() // return 1

Note

min operation will never be called if there is no number in the stack.
注意在push()里的条件，minstack.peek() >= number，保证最小值在minstack中。
第二点：

if (!stack.isEmpty()) {
                int element = stack.peek();
                if (minstack.peek() == element) {
                    minstack.pop();
                }
            }

.parseInt()，用.equals()都不对。
.intValue()，用==都是对的。

if (minstack.peek().intValue() == stack.peek().intValue()) 正确

valueOf(String) returns a new java.lang.Integer, which is the object representative of the integer,
whereas parseInt(String) returns a primitive integer type int.
intValue is an instance method whereby parseInt is a static method.
Moreover, Integer.parseInt(s) can take primitive datatype as well.

Solution

public class MinStack {
    Stack<Integer> stack = new Stack<>();
    Stack<Integer> minStack = new Stack<>();
    /** initialize your data structure here. */
    public MinStack() {
        
    }
    
    public void push(int x) {
        stack.push(x);
        if (minStack.isEmpty() || minStack.peek() >= x) minStack.push(x);
    }
    
    public void pop() {
        if (minStack.peek().equals(stack.peek())) minStack.pop();
        stack.pop();
    }
    //Better check if it is empty
    public int top() {
        return stack.peek();
    }
    //Better check if it is empty
    public int getMin() {
        return minStack.peek();
    }
}

MAX STACK

public void push(int x) {
    if (maxStack.isEmpty() || maxStack.peek() <= x) maxStack.push(x);
    stack.push(x);
}
public int pop() {
    if (maxStack.peek().equals(stack.peek())) maxStack.pop();
    return stack.pop();
}
public int max() {
    return maxStack.peek();
}