Leetcode 78. Subsets & 90. Subsets II

题目来源：leetcode

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

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,3], a solution is:

思路：http://www.cnblogs.com/felixfang/p/3775712.html

代码：

class Solution {   public List<List<Integer>> subsets(int[] nums) {                List<List<Integer>> ls=new ArrayList<List<Integer>>();//用来存放结果        if (nums.length==0)            return ls;        ls.add(new ArrayList<Integer>());//初始加入空        for (int j=0;j<nums.length;j++){            List<List<Integer>> temp=new ArrayList<>();//存放数组每增加一个元素，结果需要增加的元素            //增加的元素为；上次所有结果集中，每个list中加入当前数组的数，别忘了在加上结果集中原有的元素            for(List<Integer> a:ls){                List<Integer> merge=new ArrayList<Integer>(a);                merge.add(nums[j]);                temp.add(merge);            }            ls.addAll(temp);                        }                return ls;    }}

II    

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:

思路：存在重复元素，思路同上，注意先对数组进行排序

class Solution {    public List<List<Integer>> subsetsWithDup(int[] nums) {        List<List<Integer>> res=new ArrayList<>();        int index=1;        //List<Integer> lin=new ArrayList<>();        if(nums.length<=0)            return res;        res.add(new ArrayList<Integer>());        Arrays.sort(nums);//先进行排序        List<Integer> temp1=new ArrayList<>();                temp1.add(nums[0]);                res.add(temp1);            if(nums.length==1){                                              return res;            }                      for(int i=1;i<nums.length;i++){                   List<List<Integer>> temp=new ArrayList<>();//存放本次增加的元素                    if(nums[i]!=nums[i-1])                    {   index=res.size();                        for(List<Integer> ls:res){                            List<Integer> a=new ArrayList<>(ls);                            a.add(nums[i]);                            temp.add(a);                        }                    }                    else{                        for(int j=index;j<res.size();j++){                            List<Integer> a=new ArrayList<>(res.get(j));                            a.add(nums[i]);                            temp.add(a);                        }                                                    index=res.size();                    }                    res.addAll(temp);        }        return res;            }}