1710. Maximum Units on a Truck

You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes, where boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]:

• numberOfBoxesi is the number of boxes of type i.
• numberOfUnitsPerBoxi is the number of units in each box of the type i.

You are also given an integer truckSize, which is the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize.

Return the maximum total number of units that can be put on the truck.

Example 1:

Input: boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4
Output: 8
Explanation: There are:
- 1 box of the first type that contains 3 units.
- 2 boxes of the second type that contain 2 units each.
- 3 boxes of the third type that contain 1 unit each.
You can take all the boxes of the first and second types, and one box of the third type.
The total number of units will be = (1 * 3) + (2 * 2) + (1 * 1) = 8.

Example 2:

Input: boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10
Output: 91

Constraints:

• 1 <= boxTypes.length <= 1000
• 1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000
• 1 <= truckSize <= 106

class Solution {
    public int maximumUnits(int[][] boxTypes, int truckSize) {
        Arrays.sort(boxTypes, (a, b) -> b[1] - a[1]);
        int res = 0;
        int ind = 0;
        while(truckSize > 0 && ind < boxTypes.length) {
            if(truckSize >= boxTypes[ind][0]) {
                    truckSize -= boxTypes[ind][0];
                    res += boxTypes[ind][0] * boxTypes[ind][1];
                    ind++;
                }
            else {                
                res += boxTypes[ind][1] * truckSize;
                break;
            }
        }
            return res;
        
    }
}

greedy和sort，一开始想成背包了结果发现不用，虽然要的也是能装下的最大值。首先对数组排序，按照第二位从大到小排，因为要是最大的。

然后根据trucksize循环，如果size比当前的数量多，就加上所有的，否则就加上最后剩下的，注意有可能boxtypes的数量比trucksize小。