lintcode-medium-Subsets II

Posted 2020-07-06 哥布林工程师

Given a list of numbers that may has duplicate numbers, return all possible subsets

 

Notice

• Each element in a subset must be in non-descending order.
• The ordering between two subsets is free.
• The solution set must not contain duplicate subsets.

Example

If S = [1,2,2], a solution is:

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

Challenge

Can you do it in both recursively and iteratively?

 

class Solution {
    /**
     * @param S: A set of numbers.
     * @return: A list of lists. All valid subsets.
     */
    public ArrayList<ArrayList<Integer>> subsetsWithDup(ArrayList<Integer> S) {
        // write your code here
        
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if(S == null || S.size() == 0){
            res.add(new ArrayList<Integer>());
            return res;
        }
        
        Collections.sort(S);
        boolean[] visited = new boolean[S.size()];
        ArrayList<Integer> line = new ArrayList<Integer>();
        
        for(int i = 0; i <= S.size(); i++)
            helper(res, line, i, 0, S, visited);
        
        return res;
    }
    
    public void helper(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> line, int len, int start, ArrayList<Integer> nums, boolean[] visited){
        
        if(line.size() == len){
            res.add(new ArrayList<Integer>(line));
            return;
        }
        
        for(int i = start; i < nums.size(); i++){
            if(i > 0 && nums.get(i) == nums.get(i - 1) && !visited[i - 1])
                continue;
            
            line.add(nums.get(i));
            visited[i] = true;
            helper(res, line, len, i + 1, nums, visited);
            visited[i] = false;
            line.remove(line.size() - 1);
        }
        
        return;
    }
    
}