## Thursday, December 12, 2013

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 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