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.
Example:

Input: [1,2,2]
Output:
[[2],[1],[1,2,2],[2,2],[1,2],[]
]

难度：medium

题目：给定一可能含有重复元素的集合，返回所有可能的子集合。
注意：答案不能含有重复子集。

思路：递归，排序去重

Runtime: 3 ms, faster than 45.31% of Java online submissions for Subsets II.
Memory Usage: 35.7 MB, less than 0.97% of Java online submissions for Subsets II.

class Solution {public List<List<Integer>> subsetsWithDup(int[] nums) {List<List<Integer>> result = new ArrayList<>();Arrays.sort(nums);for (int i = 0; i < nums.length; i++) {subsetsWithDup(nums, 0, i + 1, new Stack<Integer>(), result);}result.add(new ArrayList<>());return result;}public void subsetsWithDup(int[] nums, int idx, int count, Stack<Integer> stack, List<List<Integer>> result) {if (count <= 0) {result.add(new ArrayList<>(stack));return;}for (int i = idx; i < nums.length - count + 1; i++) {if (i == idx || nums[i - 1] != nums[i]) {stack.push(nums[i]);subsetsWithDup(nums, i + 1, count - 1, stack, result);stack.pop();}}}
}