Leetcode47. Permutations II

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> ans = new ArrayList<>();
        if(nums==null||nums.length==0) return ans;
        boolean[] used = new boolean[nums.length];
        helper(ans,new ArrayList<>(),nums,used);
        return ans;
    }
    private void helper(List<List<Integer>> ans,List<Integer> tempList,int[] nums,boolean[] used){
        if(tempList.size()==nums.length&&!ans.contains(tempList)) ans.add(new ArrayList<>(tempList));
        else{
            for(int i=0;i<nums.length;i++){
                if(used[i]) continue;
                tempList.add(nums[i]);
                used[i] = true;
                helper(ans,tempList,nums,used);
                tempList.remove(tempList.size()-1);
                used[i]=false;
            }
        }
    }
}

320ms,5%。极慢…应该差在第10行的去重了。

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

    private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, boolean [] used){
        if(tempList.size() == nums.length){
            list.add(new ArrayList<>(tempList));
        } else{
            for(int i = 0; i < nums.length; i++){
                if(used[i] || i > 0 && nums[i] == nums[i-1] && !used[i - 1]) continue; 
                /*Upper Line:with ordered nums remove duplicates.
                i>0: this element is not the first in this loop
                nums[i] == nums[i-1]: duplicate situation
                !used[i-1]: nums[i-1] is not currently used.If nums[i-1] is currently used then it is not a duplicate situation.*/
                used[i] = true; 
                tempList.add(nums[i]);
                backtrack(list, tempList, nums, used);
                used[i] = false; 
                tempList.remove(tempList.size() - 1);
            }
        }
    }
}

5ms,53.9%。够了。

想了想，用fre数组代替tempList不一定快太多，因为最后还是要建一个tempList add到ans中去，节省的时间只有试错时remove的时间。所以backtracking问题都用这个套路挺好的。

此解法好像有误。