Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
getMin() -- Retrieve the minimum element in the stack.

思路：實(shí)現(xiàn)一個(gè)能隨時(shí)返回當(dāng)前棧中最小值的棧，借助一個(gè)輔助棧，這個(gè)棧負(fù)責(zé)存儲(chǔ)最小值，當(dāng)有新元素入棧時(shí)，判斷新元素和輔助棧頂?shù)拇笮￡P(guān)系，確定輔助棧入棧元素。

private Stack<Integer> stack;
private Stack<Integer> minStack;

public MinStack() {
    this.stack = new Stack<>();
    this.minStack = new Stack<>();
}

public void push(int x) {
    //stack.push(x);//bug 當(dāng)push第一個(gè)元素的時(shí)候，如果stackpush在前，第20行，minStack還是empty，peek()會(huì)報(bào)錯(cuò)
    if (stack.empty()) {
        minStack.push(x);
    } else {
        minStack.push(Math.min(x, minStack.peek()));
    }
    stack.push(x);
}

public void pop() {
    stack.pop();
    minStack.pop();
}

public int top() {
    return stack.peek();
}

public int getMin() {
    return minStack.peek();
}