LeetCode——Subsets II

最新推荐文章于 2025-12-10 14:25:37 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。该算法通过递归搜索生成所有子集，并确保结果中不含重复的子集。文章提供了一个C++实现示例，演示如何对输入进行排序并跳过重复元素来避免重复子集。

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],
  []
]

» Solve this problem

class Solution {
private:
    void search(vector<vector<int> > &ans, vector<int> &temp, vector<int> &S, int idx) {
        int x = -1000000;
        for (int i = idx + 1; i < S.size(); i++) {
            if (x == S[i]) {
                continue;
            }
            x = S[i];
            temp.push_back(x);
            ans.push_back(temp);
            search(ans, temp, S, i);
            temp.pop_back();
        }
    }
    
public:
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > ans;
        vector<int> temp;
        ans.push_back(temp);
        sort(S.begin(), S.end());
        search(ans, temp, S, -1);
        return ans;
    }
};