IT company 100-design stack with min function

Problem Description:

Defines the data structure of the stack, requiring the addition of a min function to get the smallest element of the stack.

The time complexity required for the function min, push, and Pop is O (1).

Double Space Implementation:

Save 2 stacks, each of which is the element and the current minimum value.

Compact Space Implementation:

？

Code implementation:

package oschina.mianshi;/** *  @project: oschina *  @filename: it2.java  *  @version: 0.10 *  @author: jm han *  @date:  17:08 2015/10/20  *  @comment: test purpose *  @result:  */import java.util.stack;import  static tool.util.*;class MinStack{   Stack<Integer> stacks,  Mins;   minstack () {      stacks = new stack< Integer> ();      mins   = new stack<integer> () ;    }   void push (integer x) {       Stacks.push (x);       //<=  instead of < is for repeating elements        if (Mins.isempty ()  | |  x <= mins.peek ())          mins.push (x);   }   void pop () {      integer x =  stacks.pop ();       if (X == mins.peek ())           mins.pop ();    }   integer peek () {       return stacks.peek ();    }   integer getmin () {       return mins.peek ();    }}public class it2  {   public static void main (String[] args)  {       minstack minstack = new minstack ();       Minstack.push (6);       minstack.push (2);       Minstack.push (3);       minstack.push (1);       Minstack.push (1);  &nbsP;    minstack.push (5);       minstack.push (8);       system.out.println ("min value: "  + minstack.getmin ());       minstack.pop ();       minstack.pop ();       minstack.pop ();       minstack.pop ();       system.out.println ("min value: "  + minstack.getmin ());    }}

Output:

Min Value:1min value:2

IT company 100-design stack with min function