描述
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
Note:
1 <= boxTypes.length <= 1000
1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000
1 <= truckSize <= 106
解析
根据题意，找出可以放到货车上面的最大货物单位，只需要将 boxTypes 按照第二维度从高到低进行排序，然后依次将遍历所有的 boxTypes ，每次遍历如果货车容量大于 boxTypes[i][0]，将该货物单位都加入 res，如果货车容量小于 boxTypes[i][0]，则只是将 truckSize * boxTypes[i][1] 加入 res 中，返回结果即可。
解答
class Solution(object):
def maximumUnits(self, boxTypes, truckSize):
"""
:type boxTypes: List[List[int]]
:type truckSize: int
:rtype: int
"""
boxTypes = sorted(boxTypes, key=lambda x: x[1], reverse=True)
res = 0
for i in range(len(boxTypes)):
if boxTypes[i][0] <= truckSize:
res += boxTypes[i][0] * boxTypes[i][1]
truckSize -= boxTypes[i][0]
else:
res += truckSize * boxTypes[i][1]
break
return res
运行结果
Runtime: 124 ms, faster than 86.16% of Python online submissions for Maximum Units on a Truck.
Memory Usage: 14 MB, less than 62.71% of Python online submissions for Maximum Units on a Truck.
原题链接：leetcode.com/problems/ma…
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