* 491.递增子序列

class Solution {
    List<List<Integer>>result = new ArrayList<>();
    LinkedList<Integer>path = new LinkedList<>();
    boolean[] used;
    public List<List<Integer>> findSubsequences(int[] nums) {
        //Arrays.sort(nums);
        used = new boolean[nums.length];
        backTracking(nums, 0, used);
        return result;
    }

    public void backTracking(int[] nums, int startIndex, boolean[] used){
        if(path.size() >= 2){
            result.add(new ArrayList<>(path));
        }
        if(startIndex >= nums.length) return;
        HashSet<Integer>set = new HashSet<>();
        for(int i = startIndex ; i < nums.length ; i++){
            if(set.contains(nums[i] ) == true){
                continue;
            }
            if(!path.isEmpty() && nums[i] < path.getLast()){
                continue;
            }
            set.add(nums[i]);
            used[i] = true;
            path.add(nums[i]);
            backTracking(nums, i+1, used);
            path.removeLast();
            used[i] = false;            
        }
    }
}

思路：与子集问题类似，但是不能进行排序，因为需要挑出递增数列。进行树层去重和树枝去重，树枝去重需要比较path的最后一个和当前要加入的大小

* 46.全排列

class Solution {

    List<List<Integer>> result = new ArrayList<>();// 存放符合条件结果的集合
    LinkedList<Integer> path = new LinkedList<>();// 用来存放符合条件结果
    boolean[] used;
    public List<List<Integer>> permute(int[] nums) {
        if (nums.length == 0){
            return result;
        }
        used = new boolean[nums.length];
        permuteHelper(nums);
        return result;
    }

    private void permuteHelper(int[] nums){
        if (path.size() == nums.length){
            result.add(new ArrayList<>(path));
            return;
        }
        for (int i = 0; i < nums.length; i++){
            if (used[i]){
                continue;
            }
            used[i] = true;
            path.add(nums[i]);
            permuteHelper(nums);
            path.removeLast();
            used[i] = false;
        }
    }
}

思路：全排列问题相对简单，就是进行树枝去重，i每次从0开始，只要used[i] == true，就跳过。

// 解法2：通过判断path中是否存在数字，排除已经选择的数字
class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> path = new LinkedList<>();
    public List<List<Integer>> permute(int[] nums) {
        if (nums.length == 0) return result;
        backtrack(nums, path);
        return result;
    }
    public void backtrack(int[] nums, LinkedList<Integer> path) {
        if (path.size() == nums.length) {
            result.add(new ArrayList<>(path));
        }
        for (int i =0; i < nums.length; i++) {
            // 如果path中已有，则跳过
            if (path.contains(nums[i])) {
                continue;
            } 
            path.add(nums[i]);
            backtrack(nums, path);
            path.removeLast();
        }
    }
}

解法二：通过判断path是否存在数字，排除已经选择的数字。

* 47.全排列 II

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer>path = new LinkedList<>();
    boolean[] used ;
    public List<List<Integer>> permuteUnique(int[] nums) {
        Arrays.sort(nums);
        used = new boolean[nums.length];
        Arrays.fill(used, false);
        backTracking(nums, used);
        return result;
    }

    public void backTracking(int[] nums, boolean[] used){
        if(path.size() == nums.length){
            result.add(new ArrayList<>(path));
            return;
        }
        for(int i = 0 ; i < nums.length ; i++){
            if(used[i] == true){
                continue;
            }
            if(i > 0 && nums[i] == nums[i-1] && used[i-1] == false){
                continue;
            }
            path.add(nums[i]);
            used[i] = true;
            backTracking(nums, used);
            used[i] = false;
            path.removeLast();

        }
    }
}

思路：在全排列的基础上增加了树层去重，先对nums数组进行排序，然后根据used进行树层去重，并且和全排列一样进行树枝去重。