输入:正整数数组costs、正数数组profits、正数K、正数M costs[i] 标识 i 号项目的花费 profits[i] 标识 i 号项目在扣除花费之后还能挣到的钱(利润) K 表示你只能串行的最多做 K 个项目 M 表示你初始的资金 说明:每做完一个项目,马上获得的收益,可以支持你去做下一个项目。不能并行的做项目 输出:你最后获得的最大钱数 @author wen
| 15 | * @author wen |
| 16 | */ |
| 17 | public class IPO { |
| 18 | public static class Program { |
| 19 | public int cost; |
| 20 | public int profits; |
| 21 | public Program(int cost, int profits) { |
| 22 | this.cost = cost; |
| 23 | this.profits = profits; |
| 24 | } |
| 25 | } |
| 26 | |
| 27 | public static int findMaximizedCapital(int k, int w, int[] costs, int[] profits) { |
| 28 | PriorityQueue<Program> minCostQueue = new PriorityQueue<>(new MinCostComparator()); |
| 29 | PriorityQueue<Program> maxProfitsQueue = new PriorityQueue<>(new MaxProfitsComparator()); |
| 30 | for (int i = 0; i < costs.length; i++) { |
| 31 | minCostQueue.add(new Program(costs[i], profits[i])); |
| 32 | } |
| 33 | for (int i = 0; i < k; i++) { |
| 34 | while (!minCostQueue.isEmpty() && minCostQueue.peek().cost <= w) { |
| 35 | maxProfitsQueue.add(minCostQueue.poll()); |
| 36 | } |
| 37 | if (minCostQueue.isEmpty()) { |
| 38 | return w; |
| 39 | } |
| 40 | w += maxProfitsQueue.poll().profits; |
| 41 | } |
| 42 | return w; |
| 43 | } |
| 44 | |
| 45 | public static class MinCostComparator implements Comparator<Program> { |
| 46 | @Override |
| 47 | public int compare(Program o1, Program o2) { |
| 48 | return o1.cost - o2.cost; |
| 49 | } |
| 50 | } |
| 51 | |
| 52 | public static class MaxProfitsComparator implements Comparator<Program> { |
| 53 | @Override |
| 54 | public int compare(Program o1, Program o2) { |
| 55 | return o2.profits - o1.profits; |
| 56 | } |
| 57 | } |
| 58 | } |
nothing calls this directly
no outgoing calls
no test coverage detected