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Given a set of distinct integers, nums, return all possible subsets.
Note:
For example,
If nums = [1,2,3], a solution is:
[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
Because it‘s required that subset must be non-descending order, we can sort input array first. We can write out the solution space tree.
For example, input is [1,2,3]
[]
/ | \
[1] [2] [3]
/ \ |
[1,2] [1,3] [2,3]
If element of last level is [i, j, .., k], then we could traverse elements from [k + 1, .., last] to add one element to construct complete answers.
BFS can be easily implemented to solve this problem. Time complexity is size of solution space tree, ie, Cn0 + Cn1 + Cn2 + .. + Cnn/2
1 public class Solution { 2 public List<List<Integer>> subsets(int[] nums) { 3 Arrays.sort(nums); 4 int length = nums.length; 5 List<List<Integer>> results = new ArrayList<List<Integer>>(); 6 List<List<Integer>> current = new ArrayList<List<Integer>>(); 7 results.add(new ArrayList<Integer>()); 8 Map<Integer, Integer> map = new HashMap<Integer, Integer>(); 9 for (int i = 0; i < length; i++) { 10 map.put(nums[i], i); 11 List<Integer> tmp = new ArrayList<Integer>(); 12 tmp.add(nums[i]); 13 current.add(tmp); 14 results.add(tmp); 15 } 16 while (current.size() > 0) { 17 List<List<Integer>> next = new ArrayList<List<Integer>>(); 18 int l = current.size(); 19 for (int i = 0; i < l; i++) { 20 List<Integer> currentList = current.get(i); 21 int ll = currentList.size(); 22 int last = currentList.get(ll - 1); 23 int index = map.get(last); 24 for (int j = index + 1; j < length; j++) { 25 // Note: create a new object 26 List<Integer> newList = new ArrayList<Integer>(currentList); 27 newList.add(nums[j]); 28 next.add(newList); 29 results.add(newList); 30 } 31 } 32 current = next; 33 } 34 return results; 35 } 36 }
We can transfer this problem to be list all combinations from number 0 to nums.length
And we can get combinations of each size by DFS. Same time complexity as solution 1.
1 public class Solution { 2 public List<List<Integer>> subsets(int[] nums) { 3 Arrays.sort(nums); 4 int length = nums.length; 5 List<List<Integer>> results = new ArrayList<List<Integer>>(); 6 results.add(new ArrayList<Integer>()); 7 for (int i = 1; i <= length; i++) { 8 dfs(nums, 0, results, new ArrayList<Integer>(), i); 9 } 10 return results; 11 } 12 13 private void dfs(int[] nums, int start, List<List<Integer>> results, List<Integer> list, int level) { 14 if (list.size() == level) 15 results.add(new ArrayList<Integer>(list)); 16 for (int i = start; i < nums.length; i++) { 17 list.add(nums[i]); 18 dfs(nums, i + 1, results, list, level); 19 list.remove(list.size() - 1); 20 } 21 } 22 }
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原文地址：http://www.cnblogs.com/ireneyanglan/p/4876833.html
