设计一个支持 push，pop，top 操作，并能在常数时间内检索到最小元素的栈。
push(x) – 将元素 x 推入栈中。 pop() – 删除栈顶的元素。 top() – 获取栈顶元素。 getMin() – 检索栈中的最小元素。 示例:
代码语言：javascript复制MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.getMin(); --> 返回 -2.解：
代码语言：javascript复制class MinStack {
/** initialize your data structure here. */
private Stack<Integer> stack;
private Stack<Integer> minStack;
public MinStack() {
stack = new Stack<Integer>();
minStack = new Stack<Integer>();
}
public void push(int number) {
stack.push(number);
if (minStack.empty() || minStack.peek() >= number) {
minStack.push(number);
}
}
public int pop() {
if (stack.peek().equals(minStack.peek())) {
minStack.pop();
}
return stack.pop();
}
public int top() {
return stack.peek();
}
public int getMin() {
return minStack.peek();
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
