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);
    }
    
}