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.

设计一个最小栈，使得返回栈中的最小数的这个操作的时间复杂度为常数，用双栈实现，一个栈作为最小数的保存栈，代码如下：

1 class MinStack { 2 public: 3     void push(int x)  4     { 5         s.push(x); 6         if(sMin.empty() || x <= sMin.top()) 7             sMin.push(x); 8     } 9 10     void pop() 11     {12         if(s.top() == sMin.top()){13             sMin.pop();14         }15         s.pop();16     }17 18     int top() 19     {20         return s.top();21     }22 23     int getMin() 24     {25         return sMin.top();26     }27 private:28     stack
     
       s;29     stack
      
        sMin;30 };