Subsets II

最新推荐文章于 2022-05-07 14:28:24 发布

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本文介绍了一种解决含重复元素集合的所有可能子集问题的算法。该算法通过递归方式生成所有子集，并确保结果中不含重复的子集。特别地，文章详细解释了如何跟踪上一轮插入的元素数量及其生成的子集数量，以便正确处理重复元素。

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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],
  []
]

Analysis:

lastNum - last number used to insert into res

lastSets - # of newly inserted sets generated by lastNum

In fact, if lastNum is equal to num[0], which is the number that will be inserted this time, only the combinations of this duplicate number with all of the sets inserted in the last round will be inserted into res as new subsets at this time.

public class Solution {
    public void subsetsWithDup(int[] num, ArrayList<ArrayList<Integer>> res, int lastNum, int lastSets) {
        if(num.length == 0) {
            res.add(new ArrayList<Integer>());
            return;
        }
        
        int nextSets = 0;       // # of sets inserted currently
        int size = res.size();
        for(int i=(num[0]==lastNum)?size-lastSets:0; i<size; i++) {
            ArrayList<Integer> tem = new ArrayList<Integer>(res.get(i));
            tem.add(num[0]);
            res.add(tem);
            nextSets++;
        }
        if(num[0] != lastNum) {
            ArrayList<Integer> cell = new ArrayList<Integer>();
            cell.add(num[0]);
            res.add(cell);
            nextSets++;
        }
        lastSets = nextSets;
        lastNum = num[0];
        int[] nextNum = Arrays.copyOfRange(num, 1, num.length);
        subsetsWithDup(nextNum, res, lastNum, lastSets);
        return;
    }
    
    public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        Arrays.sort(num);
        subsetsWithDup(num, res, Integer.MIN_VALUE, 0);
        return res;
    }
}