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# Problem 47: Unique Permutations

> <https://leetcode.com/problems/permutations-ii/>

## 思路

* 这道题最主要的就是去重复。

```java
public class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<Integer> path = new ArrayList<Integer>();
        List<List<Integer>> rst = new ArrayList(path);
        if (nums == null || nums.length == 0) {
            return rst;
        }

        boolean[] visited = new boolean[nums.length];
        Arrays.sort(nums);
        helper(nums, visited, path, rst);

        return rst;
    }

    private void helper(int[] nums, boolean[] visited, List<Integer> path, List<List<Integer>> rst) {
        if (path.size() == nums.length) {
            rst.add(new ArrayList(path));
        }

        for (int i = 0; i < nums.length; i++) {
            if (visited[i]) continue;
            path.add(nums[i]);
            visited[i] = true;
            helper(nums, visited, path, rst);
            path.remove(path.size() - 1);
            visited[i] = false;
            while (i < nums.length - 1 && nums[i] == nums[i + 1]) i++;
        }
    }
}
```

## 易错点

1. 去重复

   ```java
   while (i < nums.length - 1 && nums[i] == nums[i + 1]) i++;
   ```

   我像之前一样用了一个方法，但报错了

   ```java
   if (i > 0 && nums[i] == nums[i - 1]) {
    continue;
   }
   ```

   因为第一种方法，指针走到倒数第二个元素，第二种方法指针走到头。还是把两种统一起来，用第一种方法去重复吧。