标签：logs rem .so blog void list solution res pos
Given a collection of integers that might contain duplicates, nums, return all possible subsets. Note: The solution set must not contain duplicate subsets. For example, If nums = [1,2,2], a solution is: [ [2], [1], [1,2,2], [2,2], [1,2], [] ]
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> list = new ArrayList<Integer>();
if (nums.length == 0) {
result.add(list);
return result;
}
Arrays.sort(nums);
helper(result, list, nums, 0);
return result;
}
private void helper (List<List<Integer>> result, List<Integer> list, int[] nums, int pos) {
/* if (result.contains(list)) {
return;
}*/
result.add(new ArrayList(list));
for (int i = pos; i < nums.length; i++) {
if (i != pos && nums[i] == nums[i - 1]) {
continue;
}
list.add(nums[i]);
helper(result, list, nums, i + 1);
list.remove(list.size() - 1);
}
}
标签：logs rem .so blog void list solution res pos
原文地址：http://www.cnblogs.com/apanda009/p/7229971.html
