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Subsets

题目

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

Note:

  • Elements in a subset must be in non-descending order.
  • The solution set must not contain duplicate subsets.

For example,
If S = [1,2,3], a solution is:

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

方法一

使用DFS进行遍历。
    public void getSets(int[] S, int start, int end, List<List<Integer>> list, List<Integer> subList) {
        list.add(subList);
        for (int i = start; i < end; i++) {
            List<Integer> newSubList = new ArrayList<Integer>(subList);
            newSubList.add(S[i]);
            getSets(S, i + 1, end, list, newSubList);
        }
    }
    public List<List<Integer>> subsets(int[] S) {
        Arrays.sort(S);
        List<List<Integer>> list = new ArrayList<List<Integer>>();
        List<Integer> subList = new ArrayList<Integer>();
        getSets(S, 0, S.length, list, subList);
        return list;
    }

方法二

利用n和n-1之间的关系,n的集合等于n-1的集合加上n-1集合的每一个元素加上S[n]。
//    public List<List<Integer>> subsets(int[] S) {
//        Arrays.sort(S);
//        List<List<Integer>> list = new ArrayList<List<Integer>>();
//        List<Integer> subList0 = new ArrayList<Integer>();
//        list.add(subList0);
//        int len = S.length;
//        if (len == 0) {
//            return list;
//        }
//        List<Integer> subList1 = new ArrayList<Integer>();
//        subList1.add(S[0]);
//        list.add(subList1);
//        if (len == 1) {
//            return list;
//        }
//        for (int i = 1; i < len; i++) {
//            int value = http://www.mamicode.com/S[i];>