47. Permutations II 全排列可重复版本

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]
]

boolean数组不需要初始化，默认都是false

有重复的判断条件中，要加i > 0

!used[i - 1]，前一个元素还没用过 就找下一个元素了 不行

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
    List<List<Integer>> list = new ArrayList<>();
    boolean[] used = new boolean[nums.length];
    Arrays.sort(nums);
    backtrack(nums, new ArrayList<>(), used, list);
    return list;
}

private void backtrack(int [] nums, List<Integer> tempList, boolean[] used, List<List<Integer>> list){
    if (tempList.size() == nums.length)
        list.add(new ArrayList<>(tempList));
    
        for(int i = 0; i < nums.length; i++){        
        if(used[i] || (i > 0 && (nums[i] == nums[i - 1] && !used[i - 1]))) 
            continue;
            
        tempList.add(nums[i]);
        used[i] = true;
        backtrack(nums, tempList, used, list);
        tempList.remove(tempList.size() - 1);
        used[i] = false;
    }
    
} 
}

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