LeetCode 90. Subsets II

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本文介绍了一种解决可能包含重复元素的整数集合的所有可能子集问题的算法。该算法通过回溯法实现，并确保结果中没有重复的子集，且每个子集内的元素按非递减顺序排列。

Given a collection of integers that might contain duplicates, nums, 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 nums = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

nothing special, if we find the duplicate, we just skip to the next round backtrack.

#include <vector>
#include <iostream>
using namespace std;

void subsetsWithDup(vector<int>& nums, int start, vector< vector<int> >& res, vector<int>& path) {
    res.push_back(path);
    for(int i = start; i < nums.size(); ++i) {
        if(i > start && nums[i] == nums[i-1]) continue;
        path.push_back(nums[i]);
        subsetsWithDup(nums, i + 1, res, path);
        path.pop_back();
    }
}


// might contain duplicates.
vector< vector<int> > subsetsWithDup(vector<int>& nums) {
    if(nums.size() == 0) return {};
    sort(nums.begin(), nums.end());
    vector<int> path;
    vector< vector<int> > res;
    subsetsWithDup(nums, 0, res, path);
    return res;
}

int main(void) {
    vector<int> nums{1, 2, 2};
    vector< vector<int> > res = subsetsWithDup(nums);
    for(int i = 0; i < res.size(); ++i) {
        for(int j = 0; j < res[i].size(); ++j) {
            cout << res[i][j] << ",";
        }
        cout << endl;
    }
}