【題目】

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.

【思路】

注意sort，使得判斷臨接元素是否相鄰。

與leetcode78類似，多了一個重複數判斷條件

            if(i>flag&&nums[i-1]==nums[i])
                continue
 
;

【程式碼】

public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> ans=new ArrayList<>();
        List<Integer> tmp=new ArrayList<>();
        Arrays.sort(nums);
        fun(nums,ans,tmp,0);
        return
 
 ans;
    }
    
    public void fun(int nums[],List<List<Integer>> ans,List<Integer> tmp,int flag){
        ans.add(new ArrayList<>(tmp));
        for(int i=flag;i<nums.length;i++){
            if(i>flag&&nums[i-1]==nums[i])
                continue;
            tmp.add(nums[i]);
            fun(nums,ans,tmp,i
 
+1);
            tmp.remove(tmp.size()-1);
        }
    }
}