TOPCORDER SRM598 DIV2 BinPackingEasy

Problem is

Problem Statement
	Fox Ciel has some items. The weight of the i-th (0-based) item is item[i]. She wants to put all items into bins. The capacity of each bin is 300. She can put an arbitrary number of items into a single bin, but the total weight of items in a bin must be less than or equal to 300. You are given the int[] item. It is known that the weight of each item is between 101 and 300, inclusive. Return the minimal number of bins required to store all items.
Definition
	Class: BinPackingEasy Method: minBins Parameters: int[] Returns: int Method signature: int minBins(int[] item) (be sure your method is public)
Limits
	Time limit (s): 2.000 Memory limit (MB): 64
Constraints
-	item will contain between 1 and 50 elements, inclusive.
-	Each element of item will be between 101 and 300, inclusive.
Examples
0)
	{150, 150, 150, 150, 150} Returns: 3 You have five items and each bin can hold at most two of them. You need at least three bins.
1)
	{130, 140, 150, 160} Returns: 2 For example, you can distribute the items in the following way: Bin 1: 130, 150 Bin 2: 140, 160
2)
	{101, 101, 101, 101, 101, 101, 101, 101, 101} Returns: 5
3)
	{101, 200, 101, 101, 101, 101, 200, 101, 200} Returns: 6
4)
	{123, 145, 167, 213, 245, 267, 289, 132, 154, 176, 198} Returns: 8

and My Code is blow

public static int minBins2(int[] items){
  Arrays.sort(items);
 
  int sp = 0;
  int ep = Math.max(items.length -1, 0);
 
  int totalNum =0; 
  while(true){
   if(items[ep] > 199){
    totalNum ++;
    ep--;
   }else{
    break;
   }
  }
 
  while(ep >= sp){
   if(items[ep] + items[sp] <= 300){
    sp++;
   }
   ep--;
   totalNum++;
  }
 
  return totalNum;
 }


## Access Step ##

1. Sort Items
2. if items are bigger than 199 add totalNum
3. sum of first item and last item are bigger than 300
   last index -- and totalnum ++
.......

i think this is very easy