求解股票经纪人问题Java
时间: 2023-07-31 15:04:55 浏览: 140
股票经纪人问题是一道经典的贪心算法问题。以下是Java的实现:
```java
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
public class StockBroker {
private static class Stock {
int buyTime; // 买入时间
int sellTime; // 卖出时间
int profit; // 利润
Stock(int buy, int sell, int profit) {
this.buyTime = buy;
this.sellTime = sell;
this.profit = profit;
}
}
public static List<Stock> getBestStocks(int[] prices) {
List<Stock> stocks = new ArrayList<>();
int n = prices.length;
// 找出所有买卖时机和利润
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
int profit = prices[j] - prices[i];
stocks.add(new Stock(i, j, profit));
}
}
// 按照利润从大到小排序
Collections.sort(stocks, Comparator.comparingInt(s -> -s.profit));
// 找出不重叠的买卖时机
List<Stock> result = new ArrayList<>();
boolean[] used = new boolean[n];
for (Stock stock : stocks) {
boolean overlap = false;
for (int i = stock.buyTime; i <= stock.sellTime; i++) {
if (used[i]) {
overlap = true;
break;
}
}
if (!overlap) {
for (int i = stock.buyTime; i <= stock.sellTime; i++) {
used[i] = true;
}
result.add(stock);
}
}
return result;
}
public static void main(String[] args) {
int[] prices = {2, 5, 3, 7, 1, 4, 9, 6};
List<Stock> result = getBestStocks(prices);
for (Stock stock : result) {
System.out.printf("Buy at %d, sell at %d, profit %d%n", stock.buyTime, stock.sellTime, stock.profit);
}
}
}
```
在这个实现中,我们首先找出所有可能的买卖时机和利润,并按照利润从大到小排序。然后,我们从利润最大的股票开始,依次检查它的买卖时机是否与之前选择的股票重叠。如果没有重叠,则选择该股票,并标记其买卖时机上的天数为已使用,以便下次检查。最终我们得到的所有股票即为最佳的选择。
注意:此实现的时间复杂度为 $O(n^3)$,在实际问题中可能不太适用。可以使用更高效的算法来解决股票经纪人问题。
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