leetCode 90.Subsets II（子集II） 解题思路和方法

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],
  []
]

思路：这一题比subsets多一了一道反复，详细代码例如以下：

public class Solution {
    boolean[] b;
    Set<String> set;
    List<List<Integer>> list;
    Set<String> set1;
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        b = new boolean[nums.length];
        set = new HashSet<String>();
        list = new ArrayList<List<Integer>>();
        set1 = new HashSet<String>();

        Arrays.sort(nums);
        count(nums,"",nums.length,0);
        return list;
    }

    private void count(int[] nums,String s,int n,int j){
            //没有反复才加入
            if(set.add(s)){
               //以","切割数组
               String[] sa = s.split(",");
               List<Integer> al = new ArrayList<Integer>();
               for(int i = 0; i < sa.length; i++){
            	   if(sa[i].length() > 0){
            		   al.add(Integer.parseInt(sa[i]));
            	   }
               }
               Collections.sort(al);
               if(set1.add(al.toString()))
            	   list.add(al);
            }

        for(int i = j; i < nums.length;i++){
            if(!b[i]){
                b[i] = true;
                count(nums,s + "," + nums[i],n-1,i+1);
                b[i] = false;
            }
        }
    }

}

以下这样的写法更简洁：

public class Solution {
    List<List<Integer>> list;//结果集
    List<Integer> al;//每一项
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        list = new ArrayList<List<Integer>>();
        al = new ArrayList<Integer>();
        Arrays.sort(nums);
        //排列组合
        count(nums,al,0);
        return list;
    }

    private void count(int[] nums,List<Integer> al,int j){

    	list.add(new ArrayList<Integer>(al));//不反复的才加入

        for(int i = j; i < nums.length;i++){
        	if(i == j || nums[i] != nums[i-1]){//去除反复
        		al.add(nums[i]);//加入
                count(nums,al,i+1);
                al.remove(al.size()-1);//去除。为下一个结果做准备
        	}
        }
    }

}