java[001,002,003]如何效率最高的穷举出两位以上的所有组合方式？

Question

java 现有list1[11,33,22]
效率最高穷举出两位以上任意相连的方式，如：

[11,33]、[11,22]、[11,33,22]、[11,22,33]、
[33,22]、[33,11]、[33,11,22]、[33,22,11]、
[22,11]、[22,33]、[22,33,11]、[22,11,33]、

Answer 1

import java.util.*;

public class Test {

    // 使用递归实现
    public static void main(String[] args) {
        int[] nums = { 11, 33, 22 };
        for (int i = 2; i <= nums.length; i++) {
            combine(nums, new int[i], 0, 0);
        }
    }

    public static void combine(int[] nums, int[] temp, int start, int index) {
        if (index == temp.length) {
            permutation(temp, 0, temp.length - 1);
            return;
        }
        for (int i = start; i < nums.length; i++) {
            temp[index] = nums[i];
            combine(nums, temp, i + 1, index + 1);
        }
    }

    public static void permutation(int[] arr, int start, int end) {
        if (start == end) {
            System.out.println(Arrays.toString(arr));
        } else {
            for (int i = start; i <= end; i++) {
                swap(arr, start, i);
                permutation(arr, start + 1, end);
                swap(arr, start, i);
            }
        }
    }

    public static void swap(int[] arr, int i, int j) {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }

}

Answer 2

跑一下吧:

import java.util.*;

public class Main {
    public static void main(String[] args) {
        List<Integer> list1 = Arrays.asList(11,33,22);
        List<List<Integer>> res = enumerate(list1);
        System.out.println(res);
    }

    private static List<List<Integer>> enumerate(List<Integer> list) {
        List<List<Integer>> res = new ArrayList<>();

        if (list.size() <= 1) {
            return res;
        }

        for (int i = 0; i < list.size() - 1; i++) {
            Integer num1 = list.get(i);
            for (int j = i + 1; j < list.size(); j++) {
                Integer num2 = list.get(j);
                List<Integer> subList = list.subList(i, j + 1);
                res.add(subList);

                List<List<Integer>> subRes = enumerate(subList);
                for (List<Integer> sub : subRes) {
                    res.add(sub);
                }
            }
        }

        return res;
    }
}