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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();
            }
        }
    }
}

原文 https://segmentfault.com/a/1190000018129537

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