19.2.23 [LeetCode 90] Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: [1,2,2]
Output:
[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

题意

一个有重复元素的集合，求出所有不重复的子集

题解

class Solution {
public:
    void solve(vector<int>&nums, vector<vector<int>>&ans, vector<int>&now, int i) {
        int n = nums.size(), _i = i;
        for (; i < n; i++) {
            if (i != _i && nums[i] == nums[i - 1])continue;
            now.push_back(nums[i]);
            ans.push_back(now);
            solve(nums, ans, now, i + 1);
            now.pop_back();
        }
    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>>ans; vector<int>now;
        ans.push_back(vector<int>());
        solve(nums, ans, now, 0);
        return ans;
    }
};

循环的时候注意一下跳过重复元素即可