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：实现一个能随时返回当前栈中最小值的栈，借助一个辅助栈，这个栈负责存储最小值，当有新元素入栈时，判断新元素和辅助栈顶的大小关系，确定辅助栈入栈元素。
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 当push第一个元素的时候，如果stackpush在前，第20行，minStack还是empty，peek()会报错
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();
}