力扣算法题—090子集2

给定一个可能包含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。

说明：解集不能包含重复的子集。

示例:

输入: [1,2,2]
输出:
[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

#include"_000库函数.h"


class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int> &nums) {
        if (nums.empty()) return {};
        vector<vector<int>> res(1);
        sort(nums.begin(), nums.end());
        int size = 1, last = nums[0];
        for (int i = 0; i < nums.size(); ++i) {
            if (last != nums[i]) {//去除重复数字
                last = nums[i];
                size = res.size();
            }
            int newSize = res.size();
            for (int j = newSize - size; j < newSize; ++j) {
                res.push_back(res[j]);
                res.back().push_back(nums[i]);
            }
        }
        return res;
    }
};

//使用二叉树递归
class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int> &S) {
        if (S.empty()) return {};
        vector<vector<int>> res;
        vector<int> out;
        sort(S.begin(), S.end());
        getSubsets(S, 0, out, res);
        return res;
    }
    void getSubsets(vector<int> &S, int pos, vector<int> &out, vector<vector<int>> &res) {
        res.push_back(out);
        for (int i = pos; i < S.size(); ++i) {
            out.push_back(S[i]);
            getSubsets(S, i + 1, out, res);
            out.pop_back();
            while (i + 1 < S.size() && S[i] == S[i + 1]) ++i;//去除重复数据
        }
    }
};


void T090() {
    Solution s;
    vector<int>v = { 1,2,2 };
    vector<vector<int>> res;
    res = s.subsetsWithDup(v);
    for (auto &a : res) {
        for (auto b : a)
            cout << b << "  ";
        cout << endl;
    }
    cout << endl;
}