Leetcode: Subsets II

Given a collection of integers that might contain duplicates, S, 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 S = [1,2,2], a solution is:



[

  [2],

  [1],

  [1,2,2],

  [2,2],

  [1,2],

  []

]

与Subsets问题的唯一区别在于，一个满足条件的set加入到最终结果sets里面的时候，需要先检查sets里面是否已经存在这个set，如果重复，则不添加。

这还是一道列举所有case的NP的题，用recursion, 用arraylist

第二遍做法：需要已访问数组，当前后元素一样且前面元素并未访问，这个时候就是重复的case, continue, 

注意这里的判重复条件跟3Sum那种是不一样的，这里是recursion, 3Sum不是

 1     public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {

 2         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();

 3         ArrayList<Integer> item = new ArrayList<Integer>();

 4         if (num==null || num.length==0) return res;

 5         Arrays.sort(num);

 6         boolean[] visited = new boolean[num.length];

 7         for (int len=0; len<=num.length; len++) {

 8             helper(res, item, num, 0, len, visited);

 9         }

10         return res;

11     }

12     

13     public void helper(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> item, int[] num, int starter, int len, boolean[] visited) {

14         if (item.size() == len) {

15             res.add(new ArrayList<Integer>(item));

16             return;

17         }

18         for (int i=starter; i<num.length; i++) {

19             if (i>0 && num[i]==num[i-1] && !visited[i-1]) continue;

20             item.add(num[i]);

21             visited[i] = true;

22             helper(res, item, num, i+1, len, visited);

23             item.remove(item.size()-1);

24             visited[i] = false;

25         }

26     }

第一遍做法：依赖arrraylist.contains()函数来查重，这样做其实不是很好，组合一多，时间代价挺高的

 1 import java.util.*;

 2 

 3 public class Solution {

 4     public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {

 5         java.util.Arrays.sort(num);

 6         ArrayList<Integer> set = new ArrayList<Integer>();

 7         ArrayList<ArrayList<Integer>> sets = new ArrayList<ArrayList<Integer>>();

 8         

 9         for (int i = 0; i <= num.length; i++) {

10             helper(set, sets, num, i, 0);

11         }

12         return sets;

13     }

14     

15     public void helper(ArrayList<Integer> set, ArrayList<ArrayList<Integer>> sets, int[] num, int i, int start) {

16         if (set.size() == i) {

17             if (!sets.contains(set)) {

18                 sets.add(new ArrayList<Integer>(set));

19             }

20             return;

21         }

22         for (int k = start; k < num.length; k++) {

23             set.add(num[k]);

24             helper(set, sets, num, i, k+1);

25             set.remove(set.size() - 1);

26         }

27     }

28 }