LeetCode

题目：

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：

在subsets的代码上加一句话if (i > start && nums[i] == nums[i - 1]) continue;

package recursion;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class SubsetsII {

    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        List<Integer> record = new ArrayList<Integer>();
        Arrays.sort(nums);
        int n = nums.length;
        for (int k = 0; k <= n; ++k)
            generateRecord(res, record, nums, 0, n - 1, k);
        return res;
    }
    
    private void generateRecord(List<List<Integer>> res, List<Integer> record, int[] nums, int start, int end, int k) {
        if (k == 0) {
            res.add(record);
            return;
        }
        
        for (int i = start; i <= end - k + 1; ++i) { 
            if (i > start && nums[i] == nums[i - 1]) continue;
            List<Integer> newRecord = new ArrayList<Integer>(record);
            newRecord.add(nums[i]);
            generateRecord(res, newRecord, nums, i + 1, end, k - 1);
        }
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        SubsetsII s = new SubsetsII();
        int[] nums = { 1, 2, 2, 3 };
        List<List<Integer>> res = s.subsetsWithDup(nums);
        for (List<Integer> l : res) {
            for (int i : l) 
                System.out.print(i + "	");
            System.out.println();
        }
    }

}