78. Subsets && 90. Subsets II-优快云博客

本文链接：https://blog.youkuaiyun.com/Maggie_2015/article/details/50557549

本文介绍了一种生成给定整数数组所有可能子集的算法，包括处理不重复及重复元素的情况，并确保子集中元素按非递减顺序排列且不包含重复子集。

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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;
    }