JAVAアルゴリズム:重複する数字のないシーケンスを指定し、可能なすべての配列を返します(JAVAバージョン)

JAVAアルゴリズム:重複する数字のないシーケンスを指定し、その可能なすべての全配列を返します.

package com.bean.algorithm.basic;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.List;

public class PermuteRecursionNumber {

	private void swap(int[] array, int i, int j) {
		int temp = array[i];
		array[i] = array[j];
		array[j] = temp;
	}

	private void permuteRecursion(int[] nums, int start, int end, Integer[] nums_Integer, List> result) {
		if (start == end) {
			/*
			 * for(int i = 0; i temp = new ArrayList(Arrays.asList(nums_Integer));

			result.add(temp);
		}

		for (int i = start; i <= end; i++) {
			/*    start   i    */
			swap(nums, start, i);

			permuteRecursion(nums, start + 1, end, nums_Integer, result);

			/*    start   i    */
			swap(nums, start, i);
		}
	}

	public List> permute(int[] nums) {

		List> result = new ArrayList<>();

		Integer[] nums_Integer = new Integer[nums.length];

		permuteRecursion(nums, 0, nums.length - 1, nums_Integer, result);

		return result;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		PermuteRecursionNumber prn = new PermuteRecursionNumber();

		int[] nums = new int[] { 1, 2, 3, 4 };
		List> list = prn.permute(nums);
		Iterator> itx = list.iterator();
		while (itx.hasNext()) {
			List lt = itx.next();
			System.out.println(lt);
		}
		System.out.println();
	}

}

プログラムの実行結果:
[1, 2, 3, 4] [1, 2, 4, 3] [1, 3, 2, 4] [1, 3, 4, 2] [1, 4, 3, 2] [1, 4, 2, 3] [2, 1, 3, 4] [2, 1, 4, 3] [2, 3, 1, 4] [2, 3, 4, 1] [2, 4, 3, 1] [2, 4, 1, 3] [3, 2, 1, 4] [3, 2, 4, 1] [3, 1, 2, 4] [3, 1, 4, 2] [3, 4, 1, 2] [3, 4, 2, 1] [4, 2, 3, 1] [4, 2, 1, 3] [4, 3, 2, 1] [4, 3, 1, 2] [4, 1, 3, 2] [4, 1, 2, 3]