47. Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

Example:

Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

backtracking

和permutations不同的是可以有重复元素。先把数组排序。去重：在递归的时候判断当前元素与上一个元素是否相同，如果相同且上一个元素被访问过，跳过当前元素。

time: O(n! * n), space: O(n)

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> permutations = new ArrayList<>();
        boolean[] visited = new boolean[nums.length];
        Arrays.sort(nums);
        backtracking(nums, permutations, new ArrayList<>(), visited);
        return permutations;
    }
    
    private void backtracking(int[] nums, List<List<Integer>> permutations, List<Integer> tmp, boolean[] visited) {
        if(tmp.size() == nums.length)
            permutations.add(new ArrayList<>(tmp));
        
        for(int i = 0; i < nums.length; i++) {
            if(visited[i]) continue;
            if(i > 0 && nums[i] == nums[i-1] && visited[i-1]) continue;
            tmp.add(nums[i]);
            visited[i] = true;
            backtracking(nums, permutations, tmp, visited);
            visited[i] = false;
            tmp.remove(tmp.size() - 1);
        }
    }
}