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826. Most Profit Assigning Work

题目

We have jobs: difficulty[i] is the difficulty of the ith job, and profit[i] is the profit of the ith job.

Now we have some workers. worker[i] is the ability of the ith worker, which means that this worker can only complete a job with difficulty at most worker[i].

Every worker can be assigned at most one job, but one job can be completed multiple times.

For example, if 3 people attempt the same job that pays $1, then the total profit will be $3. If a worker cannot complete any job, his profit is $0.

What is the most profit we can make?

Example 1:

Input: difficulty = [2,4,6,8,10], profit = [10,20,30,40,50], worker = [4,5,6,7]
Output: 100 
Explanation: Workers are assigned jobs of difficulty [4,4,6,6] and they get profit of [20,20,30,30] seperately.

Note:

• 1 <= difficulty.length = profit.length <= 10000
• 1 <= worker.length <= 10000
• difficulty[i], profit[i], worker[i] are in range [1, 10^5]

题目大意

这道题考察的是滑动窗口的问题，也是排序相关的问题。

给出一组任务，每个任务都有一定的难度，每个任务也都有完成以后对应的收益(完成难的任务不一定收益最高)。有一批工人，每个人能处理的任务难度不同。要求输出这批工人完成任务以后的最大收益。

解题思路

先将任务按照难度排序，工人也按照能处理任务难度的能力排序。用一个数组记录下，每个 i 下标，当前能达到的最大收益。计算这个收益只需要从下标为 1 开始，依次比较自己和前一个的收益即可(因为排过序，难度是依次递增的)。有了这个难度依次递增，并且记录了最大收益的数组以后，就可以计算最终结果了。遍历一遍工人数组，如果工人的能力大于任务的难度，就加上这个最大收益。遍历完工人数组，最终结果就是最大收益。

Documentation

Index

• type SortByDiff
  • func (p SortByDiff) Less(i, j int) bool
• type Task
• type Tasks
  • func (p Tasks) Len() int
  • func (p Tasks) Swap(i, j int)

Types

type SortByDiff

type SortByDiff struct{ Tasks }

SortByDiff define

func (SortByDiff) Less

func (p SortByDiff) Less(i, j int) bool

Less define

type Task

type Task struct {
	Difficulty int
	Profit     int
}

Task define

type Tasks

type Tasks []Task

Tasks define

func (Tasks) Len

func (p Tasks) Len() int

Len define

func (Tasks) Swap

func (p Tasks) Swap(i, j int)

Swap define