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], []
]
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Analysis:
We first create a table of (val,count) to record how many times each number appears in the collection. We then recursive create the subset for each number.
Solution:
1 public class Solution { 2 public List<List<Integer>> subsetsWithDup(int[] num) { 3 List<List<Integer>> numList = new ArrayList<List<Integer>>(); 4 List<List<Integer>> res= new ArrayList<List<Integer>>(); 5 if (num.length==0) return res; 6 7 for (int i=0;i<num.length;i++){ 8 int val = num[i]; 9 int index = -1; 10 boolean hasItem = false; 11 for (int j=0;j<numList.size();j++) 12 if (numList.get(j).get(0)<val) 13 continue; 14 else if (numList.get(j).get(0)==val){ 15 hasItem = true; 16 index = j; 17 break; 18 } else { 19 index = j; 20 break; 21 } 22 if (hasItem) 23 numList.get(index).set(1,numList.get(index).get(1)+1); 24 else if (index!=-1){ 25 List<Integer> temp = new ArrayList<Integer>(); 26 temp.add(val); 27 temp.add(1); 28 numList.add(index,temp); 29 } else { 30 List<Integer> temp = new ArrayList<Integer>(); 31 temp.add(val); 32 temp.add(1); 33 numList.add(temp); 34 } 35 } 36 37 res = subsetsRecur(numList,0); 38 return res; 39 } 40 41 public List<List<Integer>> subsetsRecur(List<List<Integer>> numList, int curIndex){ 42 int val = numList.get(curIndex).get(0); 43 int num = numList.get(curIndex).get(1); 44 List<List<Integer>> resList = new ArrayList<List<Integer>>(); 45 if (curIndex==numList.size()-1){ 46 for (int i=0;i<=num;i++){ 47 List<Integer> subset = new ArrayList<Integer>(); 48 for (int j=0;j<i;j++) 49 subset.add(val); 50 resList.add(subset); 51 } 52 return resList; 53 } 54 55 List<List<Integer>> nextResList = subsetsRecur(numList,curIndex+1); 56 57 for (int i=0;i<=num;i++){ 58 List<Integer> subset = new ArrayList<Integer>(); 59 for (int j=0;j<i;j++) 60 subset.add(val); 61 for (int j=0;j<nextResList.size();j++){ 62 List<Integer> oneRes = new ArrayList<Integer>(); 63 oneRes.addAll(subset); 64 oneRes.addAll(nextResList.get(j)); 65 resList.add(oneRes); 66 } 67 } 68 return resList; 69 } 70 }
NOTE: To increase runtime speed, we can sort the num list and then scan it once to create the table, like this:
1 public class Solution { 2 public List<List<Integer>> subsetsWithDup(int[] num) { 3 List<List<Integer>> numList = new ArrayList<List<Integer>>(); 4 List<List<Integer>> res= new ArrayList<List<Integer>>(); 5 if (num.length==0) return res; 6 7 Arrays.sort(num); 8 int curVal = num[0]; 9 int count = 1; 10 for (int i=1;i<num.length;i++) 11 if (num[i]==curVal){ 12 count++; 13 } else { 14 List<Integer> record = new ArrayList<Integer>(); 15 record.add(curVal); 16 record.add(count); 17 numList.add(record); 18 curVal = num[i]; 19 count=1; 20 } 21 List<Integer> record = new ArrayList<Integer>(); 22 record.add(curVal); 23 record.add(count); 24 numList.add(record); 25 26 27 res = subsetsRecur(numList,0); 28 return res; 29 } 30 31 public List<List<Integer>> subsetsRecur(List<List<Integer>> numList, int curIndex){ 32 int val = numList.get(curIndex).get(0); 33 int num = numList.get(curIndex).get(1); 34 List<List<Integer>> resList = new ArrayList<List<Integer>>(); 35 if (curIndex==numList.size()-1){ 36 for (int i=0;i<=num;i++){ 37 List<Integer> subset = new ArrayList<Integer>(); 38 for (int j=0;j<i;j++) 39 subset.add(val); 40 resList.add(subset); 41 } 42 return resList; 43 } 44 45 List<List<Integer>> nextResList = subsetsRecur(numList,curIndex+1); 46 47 for (int i=0;i<=num;i++){ 48 List<Integer> subset = new ArrayList<Integer>(); 49 for (int j=0;j<i;j++) 50 subset.add(val); 51 for (int j=0;j<nextResList.size();j++){ 52 List<Integer> oneRes = new ArrayList<Integer>(); 53 oneRes.addAll(subset); 54 oneRes.addAll(nextResList.get(j)); 55 resList.add(oneRes); 56 } 57 } 58 return resList; 59 } 60 }