【剑指offer 面试题21】包含min函数的栈

思路：

　　通过增加一个辅助栈保存每个状态对应的最小值。栈实现的不完整，应该还包含empty()等常规函数。

 #include <iostream>
 #include <stack>
 using namespace std;

 template <typename T> class StackWithMin
 {
 public:
     void push(const T& value);
     void pop();
     T getMin();

 private:
     stack<T> DataStack;
     stack<T> MinStack;
 };

 template <typename T> void StackWithMin<T>::push(const T& value)
 {
     DataStack.push(value);

     if(MinStack.size() ==  || value < MinStack.top())
         MinStack.push(value);
     else
         MinStack.push(MinStack.top());
 }

 template <typename T> void StackWithMin<T>::pop()
 {
     DataStack.pop();
     MinStack.pop();
 }

 template <typename T> T StackWithMin<T>::getMin()
 {
     return MinStack.top();
 }

 int main()
 {
     StackWithMin<int> s;

     cout<<"push 5"<<endl;
     s.push();
     cout<<"Min: "<<s.getMin()<<endl;

     cout<<"push 3"<<endl;
     s.push();
     cout<<"Min: "<<s.getMin()<<endl;

     cout<<"push 1"<<endl;
     s.push();
     cout<<"Min: "<<s.getMin()<<endl;

     cout<<"pop"<<endl;
     s.pop();
     cout<<"Min: "<<s.getMin()<<endl;

     cout<<"pop"<<endl;
     s.pop();
     cout<<"Min: "<<s.getMin()<<endl;

     cout<<"push 10"<<endl;
     s.push();
     cout<<"Min: "<<s.getMin()<<endl;
 }

测试结果：

push
Min:
push
Min:
push
Min:
pop
Min:
pop
Min:
push
Min: