java 全排列去重

你可以使用递归和回溯来实现Java的全排列去重。下面是一个示例代码：

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class Permutation {
    public static List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return result;
        }
        boolean[] visited = new boolean[nums.length];
        List<Integer> permutation = new ArrayList<>();
        // 先排序，保证重复的元素都相邻
        Arrays.sort(nums);
        backtrack(nums, visited, permutation, result);
        return result;
    }

    private static void backtrack(int[] nums, boolean[] visited, List<Integer> permutation, List<List<Integer>> result) {
        if (permutation.size() == nums.length) {
            result.add(new ArrayList<>(permutation));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            // 如果当前元素已经被访问过，则跳过
            if (visited[i]) {
                continue;
            }
            // 如果当前元素与前一个元素相等，并且前一个元素未被访问过，则跳过，避免重复
            if (i > 0 && nums[i] == nums[i - 1] && !visited[i - 1]) {
                continue;
            }
            permutation.add(nums[i]);
            visited[i] = true;
            backtrack(nums, visited, permutation, result);
            visited[i] = false;
            permutation.remove(permutation.size() - 1);
        }
    }

    public static void main(String[] args) {
        int[] nums = {1, 1, 2};
        List<List<Integer>> result = permuteUnique(nums);
        for (List<Integer> permutation : result) {
            System.out.println(permutation);
        }
    }
}