Subsets II 求有重复元素的集合的子集 @LeetCode

与之前的Subsets题相比，改用HashSet来作为天然的过滤。

package Level4;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

/**
* 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],
[]
]
*
*/
public class S90 {

public static void main(String[] args) {
int[] num = {1,2,2};
System.out.println(subsetsWithDup(num));
}

// 与之前相比改用HashSet来过滤
public static ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {
int len = num.length;
Set<ArrayList<Integer>> set = new HashSet<ArrayList<Integer>>();
Arrays.sort(num);

// 利用2进制对应的bit来求所有subset，2进制中，0代表不选，1代表选择
for(int i=0; i<Math.pow(2,len); i++){
ArrayList<Integer> list = new ArrayList<Integer>();
int tmp = i;
// 对长为len的数，循环检查最右位是否为一
for(int j=0; j<len; j++){
int bit = tmp & 1;
tmp >>= 1;
if(bit == 1){
list.add(num[j]);
}
}
set.add(list);
}
return new ArrayList<ArrayList<Integer>>(set);
}

}


这道题假如不用HashSet过滤，那么就不适合用2进制或者选或者不选的方法：
这里参考了另一种解法
public class Solution {
public List<List<Integer>> subsetsWithDup(int[] num) {
Arrays.sort(num);

List<List<Integer>> ret = new ArrayList<List<Integer>>();
List<Integer> list = new ArrayList<Integer>();
rec(num, ret, list, 0);
return ret;
}

public void rec(int[] num, List<List<Integer>> ret, List<Integer> list, int pos) {
ret.add(new ArrayList<Integer>(list));

for(int i=pos; i<num.length; i++) {
if( i != pos && i-1 >= 0 && num[i] == num[i-1]) {
continue;
}
list.add(num[i]);
rec(num, ret, list, i+1);
list.remove(list.size()-1);
}
}
}


http://answer.ninechapter.com/solutions/subsets-ii/