全排列

题解：用一位数代替标记数组节省空间

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    public List<List<Integer>> permute(int[] nums) {
        List<Integer> list = new ArrayList<>();
        dfs(0,0,list,nums);
        return res;
    }
    void dfs(int u,int state,List<Integer> list,int[] nums){
        if(u == nums.length) {
            res.add(new ArrayList(list));
            return ;
        }
        for(int i=0;i<nums.length;i++){
            if((state>>i&1)==0){//将state右移i位之后与1相与实际上就是看state的第i位是否为0
                list.add(nums[i]);
                dfs(u+1,state|(1<<i),list,nums);//将state的第i位置为1
                list.remove(list.size()-1);
            }
        }
    }
}

Set暴力去重

class Solution {
    Set<List<Integer>> set = new HashSet<>();
    
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<Integer> list = new ArrayList<>();
        dfs(0,nums,list,0);
        return new ArrayList<>(set);
    }
    void dfs(int u,int[] nums,List<Integer> list,int state){
        if(u == nums.length){
            set.add(new ArrayList(list));
            return;
        }
        for(int i=0;i<nums.length;i++){
            if((state>>i&1)==0){
                list.add(nums[i]);
                dfs(u+1,nums,list,state|(1<<i));
                list.remove(list.size()-1);
            }
        }
    }
}