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：實現一個能隨時返回當前棧中最小值的棧，借助一個輔助棧，這個棧負責存儲最小值，當有新元素入棧時，判斷新元素和輔助棧頂的大小關系，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push第一個元素的時候，如果stackpush在前，第20行，minStack還是empty，peek()會報錯
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();
}