LeetCode OJ - Subsets 1 && 2

这道题的做法，一定得掌握啊！！！  elegant & beautiful & concise

下面是AC代码：

 1  /**

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

 3       * 这道题的做法应该要记住！！！！！

 4       * @param s

 5       * @return

 6       */

 7      public ArrayList<ArrayList<Integer>> subsets(int[] s){

 8          ArrayList<ArrayList<Integer>> r = new  ArrayList<ArrayList<Integer>>();

 9          Arrays.sort(s);

10          ArrayList<Integer> sub = new ArrayList<Integer>();

11          r.add(sub);

12          for(int e : s){

13              int cur_size = r.size();

14              for(int j=0;j<cur_size;j++){

15                  sub = new ArrayList<Integer>();

16                  sub.addAll(r.get(j));

17                  sub.add(e);

18                  r.add(sub);

19              }

20          }

21          return r;

22      }

23      /**

24       * Given a collection of integers that might contain duplicates, S, return all possible subsets.

25       * @param num

26       * @return

27       */

28      public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {

29          ArrayList<ArrayList<Integer>> r = new ArrayList<ArrayList<Integer>>();

30          Arrays.sort(num);

31          int last = Integer.MAX_VALUE;

32          int mark = 0;

33          ArrayList<Integer> sub = new ArrayList<Integer>();

34          r.add(sub);

35          for(int e:num){

36              int cur_size = r.size();

37              int begin = e==last? mark:0;

38              for(int i = begin;i<cur_size;i++){

39                  sub = new ArrayList<Integer>();

40                  sub.addAll(r.get(i));

41                  sub.add(e);

42                  r.add(sub);

43              }

44              last = e;

45              mark = cur_size;

46          }

47          return r;

48      }

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