标签：NPU example and output array length 不同 get 难度
这题讲的是 每个worker 只能做比他能力难度小的或者一样的工作， 每个工作创造的价值不同，为workers 们能创造的最高价值。
比如worker0 = 4, diff = {1,3,4} profit= {100,10,10} 那这个worker 显然要挑选 100这个高的profit 去做。 所以一句话：work 要挑选能力范围内，尽可能高价值的事去做。
所以要排序，难度从低到高。 worker 也得排序，然后找出自己能力范围内 max profit， 下一个人在这个 max profit 基础上继续寻找max profit.
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.
唯一的难度在于排序，要对 difficulty 和 profit 同时排序，建立一个Node, 然后调用排序函数。
class Node{ int diff; int profit; Node(int diff, int profit){ this.diff = diff; this.profit = profit; } } class Solution { public int maxProfitAssignment(int[] difficulty, int[] profit, int[] worker) { int N = difficulty.length; Node[] nodes = new Node[N]; for(int i=0; i<N; i++){ nodes[i] = new Node(difficulty[i],profit[i]) ; } Arrays.sort(nodes,(o1,o2)->(o1.diff-o2.diff)); Arrays.sort(worker); int max = 0; int index =0; int sum = 0; for(int i=0; i<worker.length; i++ ){ while(index<N && worker[i] >= nodes[index].diff){ max = Math.max(nodes[index].profit,max); index++; } sum+=max; } return sum; } }
826. Most Profit Assigning Work
标签：NPU example and output array length 不同 get 难度
原文地址：https://www.cnblogs.com/keepAC/p/11610753.html
