78. Subsets && 90. Subsets II

Given a set of distinct integers, nums, return all possible subsets.

Note:
Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,3], a solution is:

[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]

 public List<List<Integer>> subsets(int[] nums) {        List<List<Integer>> r = new ArrayList<List<Integer>>();        r.add(new ArrayList<Integer>());        if (nums == null || nums.length == 0) return r;        Arrays.sort(nums);        for (int i : nums) {            List<List<Integer>> tmp = new ArrayList<List<Integer>>( );            for(List<Integer> t : r){                List<Integer> x = new ArrayList<Integer>(t);                x.add(i);                tmp.add(x);            }            r.addAll(tmp);        }        return r;    }

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note:
Elements in a subset must be in non-descending order.
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],
[]
]
处理重复数字

public List<List<Integer>> subsetsWithDup(int[] nums) {        List<List<Integer>> r = new ArrayList<List<Integer>>();        List<List<Integer>> change = new ArrayList<List<Integer>>();   //记录前一次操作新增的部分        if(nums == null || nums.length == 0) return r;        Arrays.sort(nums);        r.add(new ArrayList<Integer>());        for(int i = 0;i < nums.length;i++){            List<List<Integer>> tmp = new ArrayList<List<Integer>>();            //重复。只改变前一次操作中新增的部分；前一次操作已经存在的部分已经增加过重复的数字了            if(i > 0 && nums[i] == nums[i-1]){                           for(List<Integer> t : change){                      List<Integer> x = new ArrayList<Integer>(t);                      x.add(nums[i]);                      tmp.add(x);                  }            }            else{                 for(List<Integer> t : r){    //没有重复                      List<Integer> x = new ArrayList<Integer>(t);                      x.add(nums[i]);                      tmp.add(x);                  }            }            r.addAll(tmp);            change = tmp;        }        return r;    }