微软面试100题-天天做-第二题

2.设计包含min函数的栈（栈）定义栈的数据结构，要求添加一个min函数，能够得到栈的最小元素。

要求函数min、push以及pop的时间复杂度都是O(1)。

package com.microsoft;

class Stack {
  private int size;
  private int top=-1;
  private int [] data;

  public Stack(int size){
    this.size=size;
    data=new int[size];
  }
  public void push(int value)throws Exception{
    if(top==size){
      throw new Exception("stack is full");
    }
    top++;
    data[top]=value;
    
  }
  
  public int pop()throws Exception{
    if(top<0){
      throw new Exception("stack is empty");
    }
    int ret=data[top];
    top--;
    return ret;
    
  }
  
  public int getTop(){
    return top;
  }
  
  public int getTopValue(){
    return data[top];
  }
  
}

public class MinStack{
  private Stack dataStack;
  private Stack minStack;
  
  public MinStack(int size){
    dataStack=new Stack(size);
    minStack=new Stack(size);
  } 
  
  public void push(int value)throws Exception{
    if(minStack.getTop()==-1||minStack.getTopValue()>=value){
      minStack.push(value);
    }
    dataStack.push(value);
  }
  
  public int pop()throws Exception{
    int ret=dataStack.pop();
    if(ret==minStack.getTopValue()){
      minStack.pop();
    }
    return ret;
  }
  
  public int min(){
    return minStack.getTopValue();
  }
  
  public static void main(String[]args)throws Exception{
    MinStack stack=new MinStack(10);
    stack.push(5);
    System.out.println("min:"+stack.min());
    stack.push(8);
    stack.push(5);
    stack.push(4);
    stack.push(9);
    System.out.println("min:"+stack.min());
    stack.pop();
    System.out.println("min:"+stack.min());
    stack.pop();
    System.out.println("min:"+stack.min());
    stack.pop();
    System.out.println("min:"+stack.min());
  }
  
}