在实现栈中原来功能的基础上，获得一个栈中最小值，要求时间复杂度 bigO(1)

思路：

　　准备两个栈 stackData stackMin

 package my_basic;

 import java.util.Stack;

 public class GetMinStack {
     Stack<Integer> stackData = new Stack<Integer>();
     Stack<Integer> stackMin = new Stack<Integer>();

     public void push(int pushInt) {
         if (stackMin.empty()) {
             stackMin.push(pushInt);
         }else if (pushInt < stackMin.peek()) {
             stackMin.push(pushInt);
         }else {
             stackMin.push(stackMin.peek());
         }
         stackData.push(pushInt);
     }

     public int pop() {
         if (stackData.empty()) {
             throw new RuntimeException("stack is empty!");
         }
         stackMin.pop();
         return stackData.pop();
     }

     public int getMin() {
         if (stackData.empty()) {
             throw new RuntimeException("stack is empty!");
         }
         return stackMin.peek();
     }

     public static void main(String[] args) {
         GetMinStack s = new GetMinStack();
         s.push(6);
         s.push(5);
         s.push(2);

         int res = s.getMin();
         System.out.println(res);

         s.push(7);

         res =  s.getMin();
         System.out.println(res);

         s.pop();
         s.pop();

         res =  s.getMin();
         System.out.println(res);
     }

 }