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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], []]
題解:實現一個遞迴的函數產生長度為level的所有子集放入結果中,然後迴圈的置這個level為0~S.length,即可產生所有的子集。
代碼如下:
1 public class Solution { 2 public void subsetsDfs(int[] s,int level,int pos,ArrayList<Integer> result,List<List<Integer>> answer){ 3 if(result.size() == level){ 4 answer.add(new ArrayList<Integer>(result)); 5 return; 6 } 7 for(int i = pos;i < s.length;i++){ 8 result.add(s[i]); 9 subsetsDfs(s, level, i+1, result, answer);10 result.remove(result.size()-1);11 }12 }13 public List<List<Integer>> subsets(int[] S) {14 Arrays.sort(S);15 List<List<Integer>> answer = new ArrayList<List<Integer>>();16 ArrayList<Integer> result = new ArrayList<Integer>();17 18 for(int i = 0;i <= S.length;i++)19 subsetsDfs(S, i, 0, result, answer);20 return answer;21 }22 }