标签：ble discuss nbsp and xpl get help length title
题目：
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], [] ]
链接：https://leetcode.com/problems/subsets-ii/#/description
4/22/2017
算法班
37%, 3ms
与第78题只有一句话的区别，就是第21行去重
1 public class Solution { 2 3 public List<List<Integer>> subsetsWithDup(int[] nums) { 4 // write your code here 5 List<List<Integer>> ret = new ArrayList<>(); 6 7 Arrays.sort(nums); 8 9 helper(ret, new ArrayList<Integer>(), nums, 0); 10 11 return ret; 12 } 13 14 private void helper(List<List<Integer>> ret, 15 ArrayList<Integer> subset, 16 int[] nums, 17 int startIndex) { 18 ret.add(new ArrayList<Integer>(subset)); 19 20 for (int i = startIndex; i < nums.length; i++) { 21 if (i > startIndex && nums[i] == nums[i - 1]) continue; 22 subset.add(nums[i]); 23 helper(ret, subset, nums, i + 1); 24 subset.remove(subset.size() - 1); 25 } 26 } 27 }
别人iterative的解释
https://discuss.leetcode.com/topic/4661/c-solution-and-explanation
更多讨论：
https://discuss.leetcode.com/category/98/subsets-ii
标签：ble discuss nbsp and xpl get help length title
原文地址：http://www.cnblogs.com/panini/p/6751418.html
