买帽子

[编程题] 买帽子

时间限制：1秒

空间限制：32768K

度度熊想去商场买一顶帽子，商场里有N顶帽子，有些帽子的价格可能相同。度度熊想买一顶价格第三便宜的帽子，问第三便宜的帽子价格是多少？ 

输入描述:

首先输入一个正整数N（N <= 50），接下来输入N个数表示每顶帽子的价格（价格均是正整数，且小于等于1000）

输出描述:

如果存在第三便宜的帽子，请输出这个价格是多少，否则输出-1

输入例子:

1010 10 10 10 20 20 30 30 40 40

输出例子:

30

import java.util.Scanner;
public class SuanFa {
public static void main(String[] args)
{
int[] price = new int[55]; 
int[] NewPrice = new int[1005];
Scanner scaner = new Scanner(System.in);
int count = scaner.nextInt();
int k = 0;
for(int j= 0;j< count;j++)
NewPrice[j] = 0;
for(int i=0;i<count;i++)
{
int p = scaner.nextInt();
NewPrice[p]++;
if(NewPrice[p] == 1)
price[k++] = p; 
}
int thirdP = getThirdPrice(price,0,k-1,k);
if(thirdP == -1)
System.out.println(-1);
else
  System.out.println(thirdP);
}

static int getThirdPrice(int[] price,int low,int high,int count)
{
if(count < 3)
  return -1;
int index = getIndex(price, low, high);
while(index != 2)
{
 if(index == 2)
break;
 else if(index > 2)
index = getIndex(price,low,index-1);
 else
index = getIndex(price,index+1,high);
}
return price[2];
}
static int getIndex(int[] price,int low,int high)
{
int pivot = price[low];
while(low < high)
{
while(low < high && price[high] >= pivot)
high--;
price[low] = price[high];
while(low < high && price[low] <= pivot)
low++;
price[high] = price[low];
}
price[low] = pivot;
return low;}
}