java - 算法是指数的，有没有办法让它不是这样？

Question

更新:根据 rrenaud 的精彩评论，我想澄清一下，尽管我正在生成所有排列，但顺序无关紧要(所以它不完全像旅行推销员问题......但仍然有点相似)。我关注我的问题的观察结果(尽管欢迎任何关于如何加速这个问题的见解)，结果是相互建立的，也许有一些方法(动态规划？)我可以用来建立在以前的结果之上而不是重复大量的计算。看看这些结果，您会发现结果只是相互叠加(这样做是为了将代码更改为 3 并在 for 循环中最大为 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
****

我仍在学习算法，所以我的知识储备有限。我基本上有一个递归，它随着我向它添加更多输入而呈指数增长，但我想知道是否有什么我可以用来让它不那样做。

这是一个递归示例(java 代码)，它基本上采用项目列表并使用 0-n 数量计算出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，然后才继续。我尽可能地对代码进行了注释，希望它对您来说很容易阅读:

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);
            }
        }
    }
}

如果将数量更改为 0-10，将项目更改为 10，则它变为(10^10，即 10000000000 循环)。当我分析代码时，它说大部分时间都花在了迭代器上，这是有道理的，所以我需要以某种方式减少它花在迭代上的时间(希望有一种方法让它每个项目迭代一次而不是每次递归) .

我的观察是初始的 for 循环似乎是必要的，因为它根据列表计算和检查值，但一旦完成，所有其他时间它循环重复相同的过程。我对所有算法都不熟悉(但目前正在阅读算法介绍)，所以我想知道是否有办法使它不呈指数增长，以便它可以处理更多变量(这是我程序的一部分，它更大，但理想情况下，我需要处理尽可能多但至少 0-100 件 100 件元素，并在一小时内完成，否则其他工作会失败。我无法用 100^100 实现这一点，所以我为想法而奋斗)。

如果有人有任何提示，或者如果他们看到任何解决类似问题的算法，那就太好了，因为我可以研究他们使用的逻辑，看看它是否适用。如果有人有直接的解决方案，那就太好了!

我希望这个问题不要太长并且有道理(我尽量详细)

Answer 1

如果你想生成所有可能的排列，这属于EXPTIME -像towers of Hanoi这样的完整类(class)，它比 NP 完全问题更难(因为它们都属于 PSPACE )，你不能期望比 100^100 更快。但可能对原始问题有一些启发，所以最好说出来。另外如果你原来的问题是这个，你可以对你的老板说:世界上没有人能做到这一点。你也可以让它并行，但除非使用非常强大的大型机，否则你没有任何机会得到结果。