更新：根据rrenaud的好评，我想澄清一下，虽然我正在产生所有排列顺序并不重要（所以它不完全像旅行推销员的问题......但仍然有点类似）。我关注我的问题的观察（虽然欢迎任何关于如何加快这一点的见解），结果是相互建立的，也许有一些方法（动态编程？）我可以用来建立先前的结果而不是重复很多计算。查看这些结果，您可以看到结果只是相互构建（为了将代码更改为3，并将for循环中的最大值设置为3）：
Grapes = 0
Strawberries = 0
Raspberries = 0
****
Grapes = 1
Strawberries = 0
Raspberries = 0
****
Grapes = 2
Strawberries = 0
Raspberries = 0
****
Grapes = 3
Strawberries = 0
Raspberries = 0
****
Grapes = 0
Strawberries = 1
Raspberries = 0
****
Grapes = 1
Strawberries = 1
Raspberries = 0
****
Grapes = 2
Strawberries = 1
Raspberries = 0
****
Grapes = 3
Strawberries = 1
Raspberries = 0
****
Grapes = 0
Strawberries = 2
Raspberries = 0
****
Grapes = 1
Strawberries = 2
Raspberries = 0
****
Grapes = 2
Strawberries = 2
Raspberries = 0
****
Grapes = 3
Strawberries = 2
Raspberries = 0
****
Grapes = 0
Strawberries = 3
Raspberries = 0
****
Grapes = 1
Strawberries = 3
Raspberries = 0
****
Grapes = 2
Strawberries = 3
Raspberries = 0
****
Grapes = 3
Strawberries = 3
Raspberries = 0
****
Grapes = 0
Strawberries = 0
Raspberries = 1
****
Grapes = 1
Strawberries = 0
Raspberries = 1
****
Grapes = 2
Strawberries = 0
Raspberries = 1
****
Grapes = 3
Strawberries = 0
Raspberries = 1
****
Grapes = 0
Strawberries = 1
Raspberries = 1
****
Grapes = 1
Strawberries = 1
Raspberries = 1
****
Grapes = 2
Strawberries = 1
Raspberries = 1
****
Grapes = 3
Strawberries = 1
Raspberries = 1
****
Grapes = 0
Strawberries = 2
Raspberries = 1
****
Grapes = 1
Strawberries = 2
Raspberries = 1
****
Grapes = 2
Strawberries = 2
Raspberries = 1
****
Grapes = 3
Strawberries = 2
Raspberries = 1
****
Grapes = 0
Strawberries = 3
Raspberries = 1
****
Grapes = 1
Strawberries = 3
Raspberries = 1
****
Grapes = 2
Strawberries = 3
Raspberries = 1
****
Grapes = 3
Strawberries = 3
Raspberries = 1
****
Grapes = 0
Strawberries = 0
Raspberries = 2
****
Grapes = 1
Strawberries = 0
Raspberries = 2
****
Grapes = 2
Strawberries = 0
Raspberries = 2
****
Grapes = 3
Strawberries = 0
Raspberries = 2
****
Grapes = 0
Strawberries = 1
Raspberries = 2
****
Grapes = 1
Strawberries = 1
Raspberries = 2
****
Grapes = 2
Strawberries = 1
Raspberries = 2
****
Grapes = 3
Strawberries = 1
Raspberries = 2
****
Grapes = 0
Strawberries = 2
Raspberries = 2
****
Grapes = 1
Strawberries = 2
Raspberries = 2
****
Grapes = 2
Strawberries = 2
Raspberries = 2
****
Grapes = 3
Strawberries = 2
Raspberries = 2
****
Grapes = 0
Strawberries = 3
Raspberries = 2
****
Grapes = 1
Strawberries = 3
Raspberries = 2
****
Grapes = 2
Strawberries = 3
Raspberries = 2
****
Grapes = 3
Strawberries = 3
Raspberries = 2
****
Grapes = 0
Strawberries = 0
Raspberries = 3
****
Grapes = 1
Strawberries = 0
Raspberries = 3
****
Grapes = 2
Strawberries = 0
Raspberries = 3
****
Grapes = 3
Strawberries = 0
Raspberries = 3
****
Grapes = 0
Strawberries = 1
Raspberries = 3
****
Grapes = 1
Strawberries = 1
Raspberries = 3
****
Grapes = 2
Strawberries = 1
Raspberries = 3
****
Grapes = 3
Strawberries = 1
Raspberries = 3
****
Grapes = 0
Strawberries = 2
Raspberries = 3
****
Grapes = 1
Strawberries = 2
Raspberries = 3
****
Grapes = 2
Strawberries = 2
Raspberries = 3
****
Grapes = 3
Strawberries = 2
Raspberries = 3
****
Grapes = 0
Strawberries = 3
Raspberries = 3
****
Grapes = 1
Strawberries = 3
Raspberries = 3
****
Grapes = 2
Strawberries = 3
Raspberries = 3
****
Grapes = 3
Strawberries = 3
Raspberries = 3
****
我还在学习算法，所以我的知识库有点受限。我基本上有一个递归，当我添加更多输入时，它会以指数方式成长，但我想知道是否有任何我可以使用它来使它不这样做。
这是一个递归的示例（java代码），它基本上取一个项目列表，并使用0-n数量计算每个项目的所有组合（为简单起见，我的代码有2个项目，每个项目的数量为0到5 ..变量可以在循环中改变）。为了使这更有趣，我有一个名为Q的变量，它执行一些随机处理并根据列表检查其值，然后才进行。我对代码进行了评论，因此我希望它能为您轻松阅读：
import java.util.Arrays;
import java.util.List;
//learning playground safe to delete
public class main {
public static void main(String[] args) {
System.out.println("Starting..");
Integer number_of_items = 2; //how many items should we test with, over 7 or 8 takes a long time MAX is 11(based on the number of names we have below)
long startTime = System.currentTimeMillis(); //start timer
Integer[] integers_temp = new Integer[number_of_items]; // create a list with exactly the number of items specified above
Arrays.fill(integers_temp, 0); // populate list with zeros
List<Integer> list_to_start = Arrays.asList(integers_temp); //set it as a list
String[] name_of_list_to_start = new String[] {"Grapes", "Strawberries", "Raspberries", "Blackberries", "Pineapples", "Oranges", "Prunes", "Pears", "cherries", "Peaches", "Apples"};
List<Integer> numbers_to_choose_from = Arrays.asList(new Integer[] {0, 1,2,3,4,5,6,7,8,9,10}); //list of numbers program can choose from(could be anything,just learning)
counter(list_to_start.size(), list_to_start, name_of_list_to_start, numbers_to_choose_from);
long endTime = System.currentTimeMillis();
System.out.println("Total execution time: " + (endTime-startTime));
}
private static void counter(int length, List<Integer> list_to_start, String[] name_of_list_to_start, List<Integer> numbers_to_choose_from) {
// If we've gone through everything then return the results
if (length == 0) {
for (int i = 0; i<list_to_start.size(); i++) {
System.out.println(name_of_list_to_start[i] + " = " + list_to_start.get(i));
}
System.out.println("****");
return;
}
//This part basically increments list_to_start and then the above part displays it.
for (int i = 0; i<=5; i++) {
int q = i +2; //do anything here..random just for example, right now just takes the number in the loop and adds by 10
//System.out.println(q); // this area is looped as many times as i^items, yet seems like after the first run work is duplicated.
if (length != 0 && numbers_to_choose_from.contains(q)) {
list_to_start.set((length-1), q);
counter((length-1), list_to_start, name_of_list_to_start, numbers_to_choose_from);
list_to_start.set((length-1), 0);
}
}
}
}
如果您将数量更改为0-10并将项目更改为10则会变为（10 ^ 10，即10000000000次循环）。当我描述代码时，它说大部分时间花在了迭代器上，这是有道理的，所以我需要以某种方式减少它花费迭代的时间（希望有一种方法让它每个项目迭代一次而不是每次递归） 。
我的观察是初始for循环似乎是必要的，因为它计算并检查列表中的值，但是一旦完成它所有其他循环重复相同的过程。我对所有的算法都非常陌生（但目前正在阅读算法的介绍），所以我想知道是否有一种方法可以使它成为无指数因此它可以处理更多变量（这是我的程序的一部分，它更大，但理想情况下，我需要尽可能多地处理，但至少0-100个数量的100个项目并在一小时内完成或其他工作失败。我无法用100 ^ 100实现这一点，所以我是努力争取创意）。
如果有人有任何提示或他们已经看到任何解决类似问题的算法，那将是很好的，因为我可以研究他们使用的逻辑并看看它是否适用。如果有人有直接解决方案，那就太好了！
我希望这个问题不会太长而且有意义（我尽量详细说明）
答案 0 :(得分：0)
如果你想要生成所有可能的排列，这属于EXPTIME - 像towers of Hanoi这样的完整类，它比NP完全问题更困难（因为它们都属于PSPACE） ，你不能指望超过100 ^ 100。但是原始问题可能有一些启发式，所以最好这么说。此外，如果你的原始问题是这样，你可以说你的老板：世界上没有人能做到这一点。你也可以使它平行，但除了使用功能非常强大的大型机之外，你没有任何机会获得结果。
答案 1 :(得分：0)
现在的代码正在迭代每个组合。对于n = 100，无法在合理的时间内运行。优化内部循环不会阻止它成为指数。
您需要一种不同的方法才能使其可行。如何执行此操作取决于您要解决的问题的详细信息。例如，可以通过动态编程，线性编程，最大流量或二进制搜索来解决一些优化问题。通过寻找近似解来解决其他问题。