90. Subsets II
Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).
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], [] ]
和上一題Subsets不同的就是這個集合裏的元素是可以重複的。
在第一題的基礎上,排序過後,相同的元素都是相鄰的,只要過濾掉相同的即可
class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> list = new ArrayList<>();
Arrays.sort(nums);
backtrack(list, new ArrayList<>(), nums, 0);
return list;
}
private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int start){
list.add(new ArrayList<>(tempList));
for(int i = start; i < nums.length; i++){
if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates
tempList.add(nums[i]);
backtrack(list, tempList, nums, i + 1);
tempList.remove(tempList.size() - 1);
}
}
}
---------------------------------------------
還是不理解,,埋個坑,以後填