*[Lintcode]Subsets 子集

Given a set of distinct integers, return all possible subsets.

Example

If S = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

分析：先排序，然后和求全排列算法一致。区别是不需要在回溯函数开始判断是否遍历结束并且加入结果集。这里每一步均加入结果集。

class Solution {
    /**
     * @param S: A set of numbers.
     * @return: A list of lists. All valid subsets.
     */
    public ArrayList<ArrayList<Integer>> subsets(int[] nums) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if (nums == null || nums.length == 0) {
            return result;
        }

        ArrayList<Integer> list = new ArrayList<Integer>();
        Arrays.sort(nums);
        backTrack(result, list, nums, 0);
        return result;
    }
    
    private void backTrack(ArrayList<ArrayList<Integer>> result,
        ArrayList<Integer> list, int[] nums, int pos) {
        result.add(new ArrayList<Integer>(list));

        for (int i = pos; i < nums.length; i++) {
            list.add(nums[i]);
            backTrack(result, list, nums, i + 1);
            list.remove(list.size() - 1);
        }
    }
}