定義棧的數據結構，請在該類型中實現一個可以獲得棧中所含最小元素的min函數（時間複雜度應爲

import java.util.Stack;

//我的認爲這種方式有問題，假設把最小值出棧，再求最小值就有問題

public class Solution

{ Stack<Integer> s=new Stack<Integer>();

Stack<Integer> ms=new Stack<Integer>();
 
 
public void push(int node)

{
    if(ms.isEmpty()||node<ms.peek())
	
        ms.push(node);
		
        s.push(node);
		
}
 
public void pop()
{
    if(ms.peek()==s.peek())
    ms.pop();
    s.pop();
}
 
public int top()
{
    return s.peek();
}
 
public int min()
{
    return ms.peek();
}

}

//我的認爲下面這種方案更靠譜

import java.util.Stack;

import java.util.Iterator;

public class Solution

{ Stack<Integer> s=new Stack<Integer>();

public void push(int node)
{
   s.push(node);
}
 
public void pop()
{ s.pop();
}
 
public int top()
{
    return s.peek();
}
 
public int min()
{
    int max = Integer.MAX_VALUE;
    Iterator it=s.iterator();
    while(it.hasNext()){
       int temp=(int)it.next();
        if(temp<max)
            max=temp;
    }
   return max;
}

}