leetcode Subsets II problem

最新推荐文章于 2021-02-27 13:36:33 发布

转载最新推荐文章于 2021-02-27 13:36:33 发布 · 803 阅读

interview questions 专栏收录该内容

22 篇文章

订阅专栏

本文介绍了一种处理包含重复元素的集合并返回所有可能子集的算法。通过递归和迭代两种方式实现了这一功能，确保子集中元素按非递减顺序排列且不包含重复子集。示例中使用 C++ 和 Java 实现。

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],
  []
]

recursive solution:

class Solution {
public:
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function            
        vector<int> path;
        vector<vector<int> > result;

        sort(S.begin(), S.end());
        sub(S, 0, path, result);
        return result;
    }

    void sub(vector<int> &s, int begin, vector<int> &path, vector<vector<int> > &result) {
        result.push_back(path);

        for (int i = begin; i < s.size(); ++i) {
            if (i != begin && s[i] == s[i - 1]) continue;

            path.push_back(s[i]);
            sub(s, i + 1, path, result);
            path.pop_back();
        }
    }
};

iterative solution:

public class Solution {
public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {
    // Start typing your Java solution below
    // DO NOT write main() function
    Arrays.sort(num);
    ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();
    ret.add(new ArrayList<Integer>());

    int start = 0;
    for(int i = 0; i < num.length; i++)
    {
        int size = ret.size();
        for(int j = start; j < size; j++)
        {
            ArrayList<Integer> sub = new ArrayList<Integer>(ret.get(j));
            sub.add(num[i]);
            ret.add(sub);
        }
        if(i < num.length - 1 && num[i + 1] == num[i])
            start = size;
        else
            start = 0;
    }

    return ret;
}}